How can adding an infinite number of rationals yield an irrational number?












93














For example how come $zeta(2)=sum_{n=1}^{infty}n^{-2}=frac{pi^2}{6}$. It seems counter intuitive that you can add numbers in $mathbb{Q}$ and get an irrational number.










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  • 88




    Every real number is the sum of countably many rational numbers.
    – Did
    Mar 4 '12 at 9:47






  • 54




    Properties that are preserved under a finite number of operations are not necessarily preserved under an infinite number of them.
    – Rahul
    Mar 4 '12 at 9:57






  • 6




    Real numbers have the property that the bounded infinite sum of positive real numbers is a real number because of completeness. This is not a property of rational numbers, so I would intuitively expect some bounded infinite sums of positive rational numbers not to be a rational number.
    – Henry
    Mar 4 '12 at 10:15






  • 4




    because real number are closure of rational numbers.
    – quartz
    Mar 4 '12 at 12:12






  • 3




    Another example to illustrate Rahul's point: the product of infinite positive reals can yield zero: $prod_{n=1}^infty 1/n = 0$ or more simply $prod_{n=1}^infty 0.5 = 0$
    – leonbloy
    Apr 30 '12 at 18:36


















93














For example how come $zeta(2)=sum_{n=1}^{infty}n^{-2}=frac{pi^2}{6}$. It seems counter intuitive that you can add numbers in $mathbb{Q}$ and get an irrational number.










share|cite|improve this question


















  • 88




    Every real number is the sum of countably many rational numbers.
    – Did
    Mar 4 '12 at 9:47






  • 54




    Properties that are preserved under a finite number of operations are not necessarily preserved under an infinite number of them.
    – Rahul
    Mar 4 '12 at 9:57






  • 6




    Real numbers have the property that the bounded infinite sum of positive real numbers is a real number because of completeness. This is not a property of rational numbers, so I would intuitively expect some bounded infinite sums of positive rational numbers not to be a rational number.
    – Henry
    Mar 4 '12 at 10:15






  • 4




    because real number are closure of rational numbers.
    – quartz
    Mar 4 '12 at 12:12






  • 3




    Another example to illustrate Rahul's point: the product of infinite positive reals can yield zero: $prod_{n=1}^infty 1/n = 0$ or more simply $prod_{n=1}^infty 0.5 = 0$
    – leonbloy
    Apr 30 '12 at 18:36
















93












93








93


16





For example how come $zeta(2)=sum_{n=1}^{infty}n^{-2}=frac{pi^2}{6}$. It seems counter intuitive that you can add numbers in $mathbb{Q}$ and get an irrational number.










share|cite|improve this question













For example how come $zeta(2)=sum_{n=1}^{infty}n^{-2}=frac{pi^2}{6}$. It seems counter intuitive that you can add numbers in $mathbb{Q}$ and get an irrational number.







number-theory






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share|cite|improve this question










asked Mar 4 '12 at 9:42









E.O.

5,01452550




5,01452550








  • 88




    Every real number is the sum of countably many rational numbers.
    – Did
    Mar 4 '12 at 9:47






  • 54




    Properties that are preserved under a finite number of operations are not necessarily preserved under an infinite number of them.
    – Rahul
    Mar 4 '12 at 9:57






  • 6




    Real numbers have the property that the bounded infinite sum of positive real numbers is a real number because of completeness. This is not a property of rational numbers, so I would intuitively expect some bounded infinite sums of positive rational numbers not to be a rational number.
    – Henry
    Mar 4 '12 at 10:15






  • 4




    because real number are closure of rational numbers.
    – quartz
    Mar 4 '12 at 12:12






  • 3




    Another example to illustrate Rahul's point: the product of infinite positive reals can yield zero: $prod_{n=1}^infty 1/n = 0$ or more simply $prod_{n=1}^infty 0.5 = 0$
    – leonbloy
    Apr 30 '12 at 18:36
















  • 88




    Every real number is the sum of countably many rational numbers.
    – Did
    Mar 4 '12 at 9:47






  • 54




    Properties that are preserved under a finite number of operations are not necessarily preserved under an infinite number of them.
    – Rahul
    Mar 4 '12 at 9:57






  • 6




    Real numbers have the property that the bounded infinite sum of positive real numbers is a real number because of completeness. This is not a property of rational numbers, so I would intuitively expect some bounded infinite sums of positive rational numbers not to be a rational number.
    – Henry
    Mar 4 '12 at 10:15






  • 4




    because real number are closure of rational numbers.
    – quartz
    Mar 4 '12 at 12:12






  • 3




    Another example to illustrate Rahul's point: the product of infinite positive reals can yield zero: $prod_{n=1}^infty 1/n = 0$ or more simply $prod_{n=1}^infty 0.5 = 0$
    – leonbloy
    Apr 30 '12 at 18:36










88




88




Every real number is the sum of countably many rational numbers.
– Did
Mar 4 '12 at 9:47




Every real number is the sum of countably many rational numbers.
– Did
Mar 4 '12 at 9:47




54




54




Properties that are preserved under a finite number of operations are not necessarily preserved under an infinite number of them.
– Rahul
Mar 4 '12 at 9:57




Properties that are preserved under a finite number of operations are not necessarily preserved under an infinite number of them.
– Rahul
Mar 4 '12 at 9:57




6




6




Real numbers have the property that the bounded infinite sum of positive real numbers is a real number because of completeness. This is not a property of rational numbers, so I would intuitively expect some bounded infinite sums of positive rational numbers not to be a rational number.
– Henry
Mar 4 '12 at 10:15




Real numbers have the property that the bounded infinite sum of positive real numbers is a real number because of completeness. This is not a property of rational numbers, so I would intuitively expect some bounded infinite sums of positive rational numbers not to be a rational number.
– Henry
Mar 4 '12 at 10:15




4




4




because real number are closure of rational numbers.
– quartz
Mar 4 '12 at 12:12




because real number are closure of rational numbers.
– quartz
Mar 4 '12 at 12:12




3




3




Another example to illustrate Rahul's point: the product of infinite positive reals can yield zero: $prod_{n=1}^infty 1/n = 0$ or more simply $prod_{n=1}^infty 0.5 = 0$
– leonbloy
Apr 30 '12 at 18:36






Another example to illustrate Rahul's point: the product of infinite positive reals can yield zero: $prod_{n=1}^infty 1/n = 0$ or more simply $prod_{n=1}^infty 0.5 = 0$
– leonbloy
Apr 30 '12 at 18:36












7 Answers
7






active

oldest

votes


















270














But for example $$pi=3+0.1+0.04+0.001+0.0005+0.00009+0.000002+cdots$$ and that surely does not seem strange to you...






share|cite|improve this answer

















  • 1




    Interesting... as a side note, you can determine the size of an infinite set by seeing if it can fit inside another infinite set. This isn't directly relevant here, but still is something important to understand. en.wikipedia.org/wiki/Aleph_number
    – JSWork
    Mar 5 '12 at 15:06








  • 45




    @JSWork, not only would I say that that isn't directly relevant here: it is quite irrelevant, in fact! :D
    – Mariano Suárez-Álvarez
    Mar 7 '12 at 4:11










  • @MarianoSuárez-Álvarez Is infinite sum of rationals always an irrational number ? Because recently I saw an example that the infinite sum of 1/2n for n from 1 to infinity equals 1, and 1 is rational.
    – Maths Survivor
    Dec 20 '17 at 23:14










  • @MathsSurvivor No. Your own comment has the proof.
    – user3658307
    May 11 at 18:37



















68














You can't add an infinite number of rational numbers. What you can do, though, is find a limit of a sequence of partial sums. So, $pi^2/6$ is the limit to infinity of the sequence $1, 1 + 1/4, 1 + 1/4 + 1/9, 1 + 1/4 + 1/9 + 1/16, ldots $. Writing it so that it looks like a sum is really just a shorthand.



In other words, $sum^infty_{i=1} cdots$ is actually kind of an abbreviation for $lim_{ntoinfty} sum^n_{i=1} cdots$.






share|cite|improve this answer

















  • 3




    Good point to observe!
    – user21436
    Mar 4 '12 at 10:25






  • 22




    @TonyK, nop, you really cannot. What you can do is to take the limit of sequences of partial sums: there is a difference between that and «summing an infinite sequence of numbers». In particular, the latter simply does not make sense.
    – Mariano Suárez-Álvarez
    Mar 5 '12 at 2:06






  • 7




    @TonyK, Dear TonyK: if you read Rudin with care you will see that it nowhere does that.
    – Mariano Suárez-Álvarez
    Mar 5 '12 at 20:05






  • 9




    Oh well. ${}{}$
    – Mariano Suárez-Álvarez
    Mar 5 '12 at 22:07






  • 24




    @TonyK Quoting from Rudin: "The number $s$ is called the sum of the series, but it should be clearly understood that $s$ is the limit of a sequence of sums and is not obtained simply by addition."
    – Pedro Tamaroff
    Apr 15 '13 at 19:31



















22














Others have demonstrated some examples that make clear why this can happen, but I wanted to point out the key mathematical concept here is "Completeness" of the metric space. A metric space is any set with "distance" defined between any two elements (in the case of $mathbb{Q}$, we would say $d(x,y) = |x-y|$). A sequence $x_i$ is "Cauchy" if late elements stop moving around very much, a necessary condition for a sequence to have a finite limit. To put it formally, ${x_i}$ is cauchy for $epsilon>0$, there is a sufficiently large $N$ so that for every $m,n>N$ we have $d(x_n,x_m)<epsilon$. A metric space is complete if all Cauchy sequences have a limit in the space. The canonical complete metric space is $mathbb R$, which is in fact the completion of $mathbb{Q}$, or the smallest complete set containing $mathbb Q$.



We think of an infinite sum as the limit of a sequence of partial sums:
$$sum_{n=1}^infty x_n = lim_{Ntoinfty}left( sum_{i=1}^Nx_n right)$$
As others have pointed out with a number of good counter-examples (my favorite of which is the decimal representation of an irrational number), $mathbb{Q}$ is not complete, therefore an infinite sum of elements of $mathbb Q$, for which partial sums are necessarily elements of $mathbb Q$, can converge to a value not in $mathbb Q$.






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    9














    It is counter-intuitive only if you are adding a "finite" number of rational numbers. Otherwise, as @Mariano implied, any irrational number consists of an infinite number of digits, and thus can be represented as a sum of rational numbers.






    share|cite|improve this answer



















    • 2




      This is my point. There is nothing magic with irrational numbers and they just use the same digits that rational numbers, so there is no surprise or counter-intuition in having irrational numbers representable as an infinite sum of rational numbers.
      – Rafid
      Mar 4 '12 at 10:41






    • 2




      @RahulNarain: In the decimal expressions of rational numbers, there is always either a terminating or an infinitely repeating string of digits. Irrational numbers have neither of these properties, but can still be expressed as an infinitely long non-repeating sequence digits. Though there may not be an explicit formula for this sequence, it can still be thought of as an infinite sum of rational numbers, as Rafid indicated.
      – Paul
      Mar 4 '12 at 15:00






    • 1




      @Paul, I was making a joke on the phrase "must have digits from 0 to 9" that appeared in the original version of the answer.
      – Rahul
      Mar 4 '12 at 22:31



















    4














    Besides that. Given a sequence of positive rational numbers such that their sum converges and such that $a_n > sum_{k = n + 1}^infty a_k$. Then choosing a $pm$ sign for each term of the sequence gives a new convergent series, each to a different number. By a countable-uncountable argument you get nonnumerable examples of that kind of series :).






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      3














      For example look at
      $$e = sum_{k=0}^{infty}{frac{1}{k!}} $$






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        2














        This has to see with the rate at which the sum/series converges to its limit and the Roth-Thue-Siegel theorem which allows you to use the rate of convergence to decide if the limit is rational or not.



        Maybe Emile (the OP) meant to ask something of this sort (please let me know, Emile): why do some (convergent, of course) infinite sums of rationals are rational and others are irrational?






        share|cite|improve this answer























        • this is a different question..
          – AIB
          Mar 6 '12 at 9:41










        • How so? The conditions/rate of convergence affect whether the limit is rational or not. Don't they?
          – AQP
          Mar 6 '12 at 16:54











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        7 Answers
        7






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        7 Answers
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        270














        But for example $$pi=3+0.1+0.04+0.001+0.0005+0.00009+0.000002+cdots$$ and that surely does not seem strange to you...






        share|cite|improve this answer

















        • 1




          Interesting... as a side note, you can determine the size of an infinite set by seeing if it can fit inside another infinite set. This isn't directly relevant here, but still is something important to understand. en.wikipedia.org/wiki/Aleph_number
          – JSWork
          Mar 5 '12 at 15:06








        • 45




          @JSWork, not only would I say that that isn't directly relevant here: it is quite irrelevant, in fact! :D
          – Mariano Suárez-Álvarez
          Mar 7 '12 at 4:11










        • @MarianoSuárez-Álvarez Is infinite sum of rationals always an irrational number ? Because recently I saw an example that the infinite sum of 1/2n for n from 1 to infinity equals 1, and 1 is rational.
          – Maths Survivor
          Dec 20 '17 at 23:14










        • @MathsSurvivor No. Your own comment has the proof.
          – user3658307
          May 11 at 18:37
















        270














        But for example $$pi=3+0.1+0.04+0.001+0.0005+0.00009+0.000002+cdots$$ and that surely does not seem strange to you...






        share|cite|improve this answer

















        • 1




          Interesting... as a side note, you can determine the size of an infinite set by seeing if it can fit inside another infinite set. This isn't directly relevant here, but still is something important to understand. en.wikipedia.org/wiki/Aleph_number
          – JSWork
          Mar 5 '12 at 15:06








        • 45




          @JSWork, not only would I say that that isn't directly relevant here: it is quite irrelevant, in fact! :D
          – Mariano Suárez-Álvarez
          Mar 7 '12 at 4:11










        • @MarianoSuárez-Álvarez Is infinite sum of rationals always an irrational number ? Because recently I saw an example that the infinite sum of 1/2n for n from 1 to infinity equals 1, and 1 is rational.
          – Maths Survivor
          Dec 20 '17 at 23:14










        • @MathsSurvivor No. Your own comment has the proof.
          – user3658307
          May 11 at 18:37














        270












        270








        270






        But for example $$pi=3+0.1+0.04+0.001+0.0005+0.00009+0.000002+cdots$$ and that surely does not seem strange to you...






        share|cite|improve this answer












        But for example $$pi=3+0.1+0.04+0.001+0.0005+0.00009+0.000002+cdots$$ and that surely does not seem strange to you...







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 4 '12 at 9:45









        Mariano Suárez-Álvarez

        110k7155280




        110k7155280








        • 1




          Interesting... as a side note, you can determine the size of an infinite set by seeing if it can fit inside another infinite set. This isn't directly relevant here, but still is something important to understand. en.wikipedia.org/wiki/Aleph_number
          – JSWork
          Mar 5 '12 at 15:06








        • 45




          @JSWork, not only would I say that that isn't directly relevant here: it is quite irrelevant, in fact! :D
          – Mariano Suárez-Álvarez
          Mar 7 '12 at 4:11










        • @MarianoSuárez-Álvarez Is infinite sum of rationals always an irrational number ? Because recently I saw an example that the infinite sum of 1/2n for n from 1 to infinity equals 1, and 1 is rational.
          – Maths Survivor
          Dec 20 '17 at 23:14










        • @MathsSurvivor No. Your own comment has the proof.
          – user3658307
          May 11 at 18:37














        • 1




          Interesting... as a side note, you can determine the size of an infinite set by seeing if it can fit inside another infinite set. This isn't directly relevant here, but still is something important to understand. en.wikipedia.org/wiki/Aleph_number
          – JSWork
          Mar 5 '12 at 15:06








        • 45




          @JSWork, not only would I say that that isn't directly relevant here: it is quite irrelevant, in fact! :D
          – Mariano Suárez-Álvarez
          Mar 7 '12 at 4:11










        • @MarianoSuárez-Álvarez Is infinite sum of rationals always an irrational number ? Because recently I saw an example that the infinite sum of 1/2n for n from 1 to infinity equals 1, and 1 is rational.
          – Maths Survivor
          Dec 20 '17 at 23:14










        • @MathsSurvivor No. Your own comment has the proof.
          – user3658307
          May 11 at 18:37








        1




        1




        Interesting... as a side note, you can determine the size of an infinite set by seeing if it can fit inside another infinite set. This isn't directly relevant here, but still is something important to understand. en.wikipedia.org/wiki/Aleph_number
        – JSWork
        Mar 5 '12 at 15:06






        Interesting... as a side note, you can determine the size of an infinite set by seeing if it can fit inside another infinite set. This isn't directly relevant here, but still is something important to understand. en.wikipedia.org/wiki/Aleph_number
        – JSWork
        Mar 5 '12 at 15:06






        45




        45




        @JSWork, not only would I say that that isn't directly relevant here: it is quite irrelevant, in fact! :D
        – Mariano Suárez-Álvarez
        Mar 7 '12 at 4:11




        @JSWork, not only would I say that that isn't directly relevant here: it is quite irrelevant, in fact! :D
        – Mariano Suárez-Álvarez
        Mar 7 '12 at 4:11












        @MarianoSuárez-Álvarez Is infinite sum of rationals always an irrational number ? Because recently I saw an example that the infinite sum of 1/2n for n from 1 to infinity equals 1, and 1 is rational.
        – Maths Survivor
        Dec 20 '17 at 23:14




        @MarianoSuárez-Álvarez Is infinite sum of rationals always an irrational number ? Because recently I saw an example that the infinite sum of 1/2n for n from 1 to infinity equals 1, and 1 is rational.
        – Maths Survivor
        Dec 20 '17 at 23:14












        @MathsSurvivor No. Your own comment has the proof.
        – user3658307
        May 11 at 18:37




        @MathsSurvivor No. Your own comment has the proof.
        – user3658307
        May 11 at 18:37











        68














        You can't add an infinite number of rational numbers. What you can do, though, is find a limit of a sequence of partial sums. So, $pi^2/6$ is the limit to infinity of the sequence $1, 1 + 1/4, 1 + 1/4 + 1/9, 1 + 1/4 + 1/9 + 1/16, ldots $. Writing it so that it looks like a sum is really just a shorthand.



        In other words, $sum^infty_{i=1} cdots$ is actually kind of an abbreviation for $lim_{ntoinfty} sum^n_{i=1} cdots$.






        share|cite|improve this answer

















        • 3




          Good point to observe!
          – user21436
          Mar 4 '12 at 10:25






        • 22




          @TonyK, nop, you really cannot. What you can do is to take the limit of sequences of partial sums: there is a difference between that and «summing an infinite sequence of numbers». In particular, the latter simply does not make sense.
          – Mariano Suárez-Álvarez
          Mar 5 '12 at 2:06






        • 7




          @TonyK, Dear TonyK: if you read Rudin with care you will see that it nowhere does that.
          – Mariano Suárez-Álvarez
          Mar 5 '12 at 20:05






        • 9




          Oh well. ${}{}$
          – Mariano Suárez-Álvarez
          Mar 5 '12 at 22:07






        • 24




          @TonyK Quoting from Rudin: "The number $s$ is called the sum of the series, but it should be clearly understood that $s$ is the limit of a sequence of sums and is not obtained simply by addition."
          – Pedro Tamaroff
          Apr 15 '13 at 19:31
















        68














        You can't add an infinite number of rational numbers. What you can do, though, is find a limit of a sequence of partial sums. So, $pi^2/6$ is the limit to infinity of the sequence $1, 1 + 1/4, 1 + 1/4 + 1/9, 1 + 1/4 + 1/9 + 1/16, ldots $. Writing it so that it looks like a sum is really just a shorthand.



        In other words, $sum^infty_{i=1} cdots$ is actually kind of an abbreviation for $lim_{ntoinfty} sum^n_{i=1} cdots$.






        share|cite|improve this answer

















        • 3




          Good point to observe!
          – user21436
          Mar 4 '12 at 10:25






        • 22




          @TonyK, nop, you really cannot. What you can do is to take the limit of sequences of partial sums: there is a difference between that and «summing an infinite sequence of numbers». In particular, the latter simply does not make sense.
          – Mariano Suárez-Álvarez
          Mar 5 '12 at 2:06






        • 7




          @TonyK, Dear TonyK: if you read Rudin with care you will see that it nowhere does that.
          – Mariano Suárez-Álvarez
          Mar 5 '12 at 20:05






        • 9




          Oh well. ${}{}$
          – Mariano Suárez-Álvarez
          Mar 5 '12 at 22:07






        • 24




          @TonyK Quoting from Rudin: "The number $s$ is called the sum of the series, but it should be clearly understood that $s$ is the limit of a sequence of sums and is not obtained simply by addition."
          – Pedro Tamaroff
          Apr 15 '13 at 19:31














        68












        68








        68






        You can't add an infinite number of rational numbers. What you can do, though, is find a limit of a sequence of partial sums. So, $pi^2/6$ is the limit to infinity of the sequence $1, 1 + 1/4, 1 + 1/4 + 1/9, 1 + 1/4 + 1/9 + 1/16, ldots $. Writing it so that it looks like a sum is really just a shorthand.



        In other words, $sum^infty_{i=1} cdots$ is actually kind of an abbreviation for $lim_{ntoinfty} sum^n_{i=1} cdots$.






        share|cite|improve this answer












        You can't add an infinite number of rational numbers. What you can do, though, is find a limit of a sequence of partial sums. So, $pi^2/6$ is the limit to infinity of the sequence $1, 1 + 1/4, 1 + 1/4 + 1/9, 1 + 1/4 + 1/9 + 1/16, ldots $. Writing it so that it looks like a sum is really just a shorthand.



        In other words, $sum^infty_{i=1} cdots$ is actually kind of an abbreviation for $lim_{ntoinfty} sum^n_{i=1} cdots$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 4 '12 at 10:20







        user22805















        • 3




          Good point to observe!
          – user21436
          Mar 4 '12 at 10:25






        • 22




          @TonyK, nop, you really cannot. What you can do is to take the limit of sequences of partial sums: there is a difference between that and «summing an infinite sequence of numbers». In particular, the latter simply does not make sense.
          – Mariano Suárez-Álvarez
          Mar 5 '12 at 2:06






        • 7




          @TonyK, Dear TonyK: if you read Rudin with care you will see that it nowhere does that.
          – Mariano Suárez-Álvarez
          Mar 5 '12 at 20:05






        • 9




          Oh well. ${}{}$
          – Mariano Suárez-Álvarez
          Mar 5 '12 at 22:07






        • 24




          @TonyK Quoting from Rudin: "The number $s$ is called the sum of the series, but it should be clearly understood that $s$ is the limit of a sequence of sums and is not obtained simply by addition."
          – Pedro Tamaroff
          Apr 15 '13 at 19:31














        • 3




          Good point to observe!
          – user21436
          Mar 4 '12 at 10:25






        • 22




          @TonyK, nop, you really cannot. What you can do is to take the limit of sequences of partial sums: there is a difference between that and «summing an infinite sequence of numbers». In particular, the latter simply does not make sense.
          – Mariano Suárez-Álvarez
          Mar 5 '12 at 2:06






        • 7




          @TonyK, Dear TonyK: if you read Rudin with care you will see that it nowhere does that.
          – Mariano Suárez-Álvarez
          Mar 5 '12 at 20:05






        • 9




          Oh well. ${}{}$
          – Mariano Suárez-Álvarez
          Mar 5 '12 at 22:07






        • 24




          @TonyK Quoting from Rudin: "The number $s$ is called the sum of the series, but it should be clearly understood that $s$ is the limit of a sequence of sums and is not obtained simply by addition."
          – Pedro Tamaroff
          Apr 15 '13 at 19:31








        3




        3




        Good point to observe!
        – user21436
        Mar 4 '12 at 10:25




        Good point to observe!
        – user21436
        Mar 4 '12 at 10:25




        22




        22




        @TonyK, nop, you really cannot. What you can do is to take the limit of sequences of partial sums: there is a difference between that and «summing an infinite sequence of numbers». In particular, the latter simply does not make sense.
        – Mariano Suárez-Álvarez
        Mar 5 '12 at 2:06




        @TonyK, nop, you really cannot. What you can do is to take the limit of sequences of partial sums: there is a difference between that and «summing an infinite sequence of numbers». In particular, the latter simply does not make sense.
        – Mariano Suárez-Álvarez
        Mar 5 '12 at 2:06




        7




        7




        @TonyK, Dear TonyK: if you read Rudin with care you will see that it nowhere does that.
        – Mariano Suárez-Álvarez
        Mar 5 '12 at 20:05




        @TonyK, Dear TonyK: if you read Rudin with care you will see that it nowhere does that.
        – Mariano Suárez-Álvarez
        Mar 5 '12 at 20:05




        9




        9




        Oh well. ${}{}$
        – Mariano Suárez-Álvarez
        Mar 5 '12 at 22:07




        Oh well. ${}{}$
        – Mariano Suárez-Álvarez
        Mar 5 '12 at 22:07




        24




        24




        @TonyK Quoting from Rudin: "The number $s$ is called the sum of the series, but it should be clearly understood that $s$ is the limit of a sequence of sums and is not obtained simply by addition."
        – Pedro Tamaroff
        Apr 15 '13 at 19:31




        @TonyK Quoting from Rudin: "The number $s$ is called the sum of the series, but it should be clearly understood that $s$ is the limit of a sequence of sums and is not obtained simply by addition."
        – Pedro Tamaroff
        Apr 15 '13 at 19:31











        22














        Others have demonstrated some examples that make clear why this can happen, but I wanted to point out the key mathematical concept here is "Completeness" of the metric space. A metric space is any set with "distance" defined between any two elements (in the case of $mathbb{Q}$, we would say $d(x,y) = |x-y|$). A sequence $x_i$ is "Cauchy" if late elements stop moving around very much, a necessary condition for a sequence to have a finite limit. To put it formally, ${x_i}$ is cauchy for $epsilon>0$, there is a sufficiently large $N$ so that for every $m,n>N$ we have $d(x_n,x_m)<epsilon$. A metric space is complete if all Cauchy sequences have a limit in the space. The canonical complete metric space is $mathbb R$, which is in fact the completion of $mathbb{Q}$, or the smallest complete set containing $mathbb Q$.



        We think of an infinite sum as the limit of a sequence of partial sums:
        $$sum_{n=1}^infty x_n = lim_{Ntoinfty}left( sum_{i=1}^Nx_n right)$$
        As others have pointed out with a number of good counter-examples (my favorite of which is the decimal representation of an irrational number), $mathbb{Q}$ is not complete, therefore an infinite sum of elements of $mathbb Q$, for which partial sums are necessarily elements of $mathbb Q$, can converge to a value not in $mathbb Q$.






        share|cite|improve this answer


























          22














          Others have demonstrated some examples that make clear why this can happen, but I wanted to point out the key mathematical concept here is "Completeness" of the metric space. A metric space is any set with "distance" defined between any two elements (in the case of $mathbb{Q}$, we would say $d(x,y) = |x-y|$). A sequence $x_i$ is "Cauchy" if late elements stop moving around very much, a necessary condition for a sequence to have a finite limit. To put it formally, ${x_i}$ is cauchy for $epsilon>0$, there is a sufficiently large $N$ so that for every $m,n>N$ we have $d(x_n,x_m)<epsilon$. A metric space is complete if all Cauchy sequences have a limit in the space. The canonical complete metric space is $mathbb R$, which is in fact the completion of $mathbb{Q}$, or the smallest complete set containing $mathbb Q$.



          We think of an infinite sum as the limit of a sequence of partial sums:
          $$sum_{n=1}^infty x_n = lim_{Ntoinfty}left( sum_{i=1}^Nx_n right)$$
          As others have pointed out with a number of good counter-examples (my favorite of which is the decimal representation of an irrational number), $mathbb{Q}$ is not complete, therefore an infinite sum of elements of $mathbb Q$, for which partial sums are necessarily elements of $mathbb Q$, can converge to a value not in $mathbb Q$.






          share|cite|improve this answer
























            22












            22








            22






            Others have demonstrated some examples that make clear why this can happen, but I wanted to point out the key mathematical concept here is "Completeness" of the metric space. A metric space is any set with "distance" defined between any two elements (in the case of $mathbb{Q}$, we would say $d(x,y) = |x-y|$). A sequence $x_i$ is "Cauchy" if late elements stop moving around very much, a necessary condition for a sequence to have a finite limit. To put it formally, ${x_i}$ is cauchy for $epsilon>0$, there is a sufficiently large $N$ so that for every $m,n>N$ we have $d(x_n,x_m)<epsilon$. A metric space is complete if all Cauchy sequences have a limit in the space. The canonical complete metric space is $mathbb R$, which is in fact the completion of $mathbb{Q}$, or the smallest complete set containing $mathbb Q$.



            We think of an infinite sum as the limit of a sequence of partial sums:
            $$sum_{n=1}^infty x_n = lim_{Ntoinfty}left( sum_{i=1}^Nx_n right)$$
            As others have pointed out with a number of good counter-examples (my favorite of which is the decimal representation of an irrational number), $mathbb{Q}$ is not complete, therefore an infinite sum of elements of $mathbb Q$, for which partial sums are necessarily elements of $mathbb Q$, can converge to a value not in $mathbb Q$.






            share|cite|improve this answer












            Others have demonstrated some examples that make clear why this can happen, but I wanted to point out the key mathematical concept here is "Completeness" of the metric space. A metric space is any set with "distance" defined between any two elements (in the case of $mathbb{Q}$, we would say $d(x,y) = |x-y|$). A sequence $x_i$ is "Cauchy" if late elements stop moving around very much, a necessary condition for a sequence to have a finite limit. To put it formally, ${x_i}$ is cauchy for $epsilon>0$, there is a sufficiently large $N$ so that for every $m,n>N$ we have $d(x_n,x_m)<epsilon$. A metric space is complete if all Cauchy sequences have a limit in the space. The canonical complete metric space is $mathbb R$, which is in fact the completion of $mathbb{Q}$, or the smallest complete set containing $mathbb Q$.



            We think of an infinite sum as the limit of a sequence of partial sums:
            $$sum_{n=1}^infty x_n = lim_{Ntoinfty}left( sum_{i=1}^Nx_n right)$$
            As others have pointed out with a number of good counter-examples (my favorite of which is the decimal representation of an irrational number), $mathbb{Q}$ is not complete, therefore an infinite sum of elements of $mathbb Q$, for which partial sums are necessarily elements of $mathbb Q$, can converge to a value not in $mathbb Q$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Mar 4 '12 at 22:17









            Neil Peterman

            32114




            32114























                9














                It is counter-intuitive only if you are adding a "finite" number of rational numbers. Otherwise, as @Mariano implied, any irrational number consists of an infinite number of digits, and thus can be represented as a sum of rational numbers.






                share|cite|improve this answer



















                • 2




                  This is my point. There is nothing magic with irrational numbers and they just use the same digits that rational numbers, so there is no surprise or counter-intuition in having irrational numbers representable as an infinite sum of rational numbers.
                  – Rafid
                  Mar 4 '12 at 10:41






                • 2




                  @RahulNarain: In the decimal expressions of rational numbers, there is always either a terminating or an infinitely repeating string of digits. Irrational numbers have neither of these properties, but can still be expressed as an infinitely long non-repeating sequence digits. Though there may not be an explicit formula for this sequence, it can still be thought of as an infinite sum of rational numbers, as Rafid indicated.
                  – Paul
                  Mar 4 '12 at 15:00






                • 1




                  @Paul, I was making a joke on the phrase "must have digits from 0 to 9" that appeared in the original version of the answer.
                  – Rahul
                  Mar 4 '12 at 22:31
















                9














                It is counter-intuitive only if you are adding a "finite" number of rational numbers. Otherwise, as @Mariano implied, any irrational number consists of an infinite number of digits, and thus can be represented as a sum of rational numbers.






                share|cite|improve this answer



















                • 2




                  This is my point. There is nothing magic with irrational numbers and they just use the same digits that rational numbers, so there is no surprise or counter-intuition in having irrational numbers representable as an infinite sum of rational numbers.
                  – Rafid
                  Mar 4 '12 at 10:41






                • 2




                  @RahulNarain: In the decimal expressions of rational numbers, there is always either a terminating or an infinitely repeating string of digits. Irrational numbers have neither of these properties, but can still be expressed as an infinitely long non-repeating sequence digits. Though there may not be an explicit formula for this sequence, it can still be thought of as an infinite sum of rational numbers, as Rafid indicated.
                  – Paul
                  Mar 4 '12 at 15:00






                • 1




                  @Paul, I was making a joke on the phrase "must have digits from 0 to 9" that appeared in the original version of the answer.
                  – Rahul
                  Mar 4 '12 at 22:31














                9












                9








                9






                It is counter-intuitive only if you are adding a "finite" number of rational numbers. Otherwise, as @Mariano implied, any irrational number consists of an infinite number of digits, and thus can be represented as a sum of rational numbers.






                share|cite|improve this answer














                It is counter-intuitive only if you are adding a "finite" number of rational numbers. Otherwise, as @Mariano implied, any irrational number consists of an infinite number of digits, and thus can be represented as a sum of rational numbers.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Mar 4 '12 at 10:44

























                answered Mar 4 '12 at 10:24









                Rafid

                6121710




                6121710








                • 2




                  This is my point. There is nothing magic with irrational numbers and they just use the same digits that rational numbers, so there is no surprise or counter-intuition in having irrational numbers representable as an infinite sum of rational numbers.
                  – Rafid
                  Mar 4 '12 at 10:41






                • 2




                  @RahulNarain: In the decimal expressions of rational numbers, there is always either a terminating or an infinitely repeating string of digits. Irrational numbers have neither of these properties, but can still be expressed as an infinitely long non-repeating sequence digits. Though there may not be an explicit formula for this sequence, it can still be thought of as an infinite sum of rational numbers, as Rafid indicated.
                  – Paul
                  Mar 4 '12 at 15:00






                • 1




                  @Paul, I was making a joke on the phrase "must have digits from 0 to 9" that appeared in the original version of the answer.
                  – Rahul
                  Mar 4 '12 at 22:31














                • 2




                  This is my point. There is nothing magic with irrational numbers and they just use the same digits that rational numbers, so there is no surprise or counter-intuition in having irrational numbers representable as an infinite sum of rational numbers.
                  – Rafid
                  Mar 4 '12 at 10:41






                • 2




                  @RahulNarain: In the decimal expressions of rational numbers, there is always either a terminating or an infinitely repeating string of digits. Irrational numbers have neither of these properties, but can still be expressed as an infinitely long non-repeating sequence digits. Though there may not be an explicit formula for this sequence, it can still be thought of as an infinite sum of rational numbers, as Rafid indicated.
                  – Paul
                  Mar 4 '12 at 15:00






                • 1




                  @Paul, I was making a joke on the phrase "must have digits from 0 to 9" that appeared in the original version of the answer.
                  – Rahul
                  Mar 4 '12 at 22:31








                2




                2




                This is my point. There is nothing magic with irrational numbers and they just use the same digits that rational numbers, so there is no surprise or counter-intuition in having irrational numbers representable as an infinite sum of rational numbers.
                – Rafid
                Mar 4 '12 at 10:41




                This is my point. There is nothing magic with irrational numbers and they just use the same digits that rational numbers, so there is no surprise or counter-intuition in having irrational numbers representable as an infinite sum of rational numbers.
                – Rafid
                Mar 4 '12 at 10:41




                2




                2




                @RahulNarain: In the decimal expressions of rational numbers, there is always either a terminating or an infinitely repeating string of digits. Irrational numbers have neither of these properties, but can still be expressed as an infinitely long non-repeating sequence digits. Though there may not be an explicit formula for this sequence, it can still be thought of as an infinite sum of rational numbers, as Rafid indicated.
                – Paul
                Mar 4 '12 at 15:00




                @RahulNarain: In the decimal expressions of rational numbers, there is always either a terminating or an infinitely repeating string of digits. Irrational numbers have neither of these properties, but can still be expressed as an infinitely long non-repeating sequence digits. Though there may not be an explicit formula for this sequence, it can still be thought of as an infinite sum of rational numbers, as Rafid indicated.
                – Paul
                Mar 4 '12 at 15:00




                1




                1




                @Paul, I was making a joke on the phrase "must have digits from 0 to 9" that appeared in the original version of the answer.
                – Rahul
                Mar 4 '12 at 22:31




                @Paul, I was making a joke on the phrase "must have digits from 0 to 9" that appeared in the original version of the answer.
                – Rahul
                Mar 4 '12 at 22:31











                4














                Besides that. Given a sequence of positive rational numbers such that their sum converges and such that $a_n > sum_{k = n + 1}^infty a_k$. Then choosing a $pm$ sign for each term of the sequence gives a new convergent series, each to a different number. By a countable-uncountable argument you get nonnumerable examples of that kind of series :).






                share|cite|improve this answer


























                  4














                  Besides that. Given a sequence of positive rational numbers such that their sum converges and such that $a_n > sum_{k = n + 1}^infty a_k$. Then choosing a $pm$ sign for each term of the sequence gives a new convergent series, each to a different number. By a countable-uncountable argument you get nonnumerable examples of that kind of series :).






                  share|cite|improve this answer
























                    4












                    4








                    4






                    Besides that. Given a sequence of positive rational numbers such that their sum converges and such that $a_n > sum_{k = n + 1}^infty a_k$. Then choosing a $pm$ sign for each term of the sequence gives a new convergent series, each to a different number. By a countable-uncountable argument you get nonnumerable examples of that kind of series :).






                    share|cite|improve this answer












                    Besides that. Given a sequence of positive rational numbers such that their sum converges and such that $a_n > sum_{k = n + 1}^infty a_k$. Then choosing a $pm$ sign for each term of the sequence gives a new convergent series, each to a different number. By a countable-uncountable argument you get nonnumerable examples of that kind of series :).







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Mar 4 '12 at 12:45









                    student

                    411




                    411























                        3














                        For example look at
                        $$e = sum_{k=0}^{infty}{frac{1}{k!}} $$






                        share|cite|improve this answer


























                          3














                          For example look at
                          $$e = sum_{k=0}^{infty}{frac{1}{k!}} $$






                          share|cite|improve this answer
























                            3












                            3








                            3






                            For example look at
                            $$e = sum_{k=0}^{infty}{frac{1}{k!}} $$






                            share|cite|improve this answer












                            For example look at
                            $$e = sum_{k=0}^{infty}{frac{1}{k!}} $$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Aug 2 '15 at 7:43







                            user210387






























                                2














                                This has to see with the rate at which the sum/series converges to its limit and the Roth-Thue-Siegel theorem which allows you to use the rate of convergence to decide if the limit is rational or not.



                                Maybe Emile (the OP) meant to ask something of this sort (please let me know, Emile): why do some (convergent, of course) infinite sums of rationals are rational and others are irrational?






                                share|cite|improve this answer























                                • this is a different question..
                                  – AIB
                                  Mar 6 '12 at 9:41










                                • How so? The conditions/rate of convergence affect whether the limit is rational or not. Don't they?
                                  – AQP
                                  Mar 6 '12 at 16:54
















                                2














                                This has to see with the rate at which the sum/series converges to its limit and the Roth-Thue-Siegel theorem which allows you to use the rate of convergence to decide if the limit is rational or not.



                                Maybe Emile (the OP) meant to ask something of this sort (please let me know, Emile): why do some (convergent, of course) infinite sums of rationals are rational and others are irrational?






                                share|cite|improve this answer























                                • this is a different question..
                                  – AIB
                                  Mar 6 '12 at 9:41










                                • How so? The conditions/rate of convergence affect whether the limit is rational or not. Don't they?
                                  – AQP
                                  Mar 6 '12 at 16:54














                                2












                                2








                                2






                                This has to see with the rate at which the sum/series converges to its limit and the Roth-Thue-Siegel theorem which allows you to use the rate of convergence to decide if the limit is rational or not.



                                Maybe Emile (the OP) meant to ask something of this sort (please let me know, Emile): why do some (convergent, of course) infinite sums of rationals are rational and others are irrational?






                                share|cite|improve this answer














                                This has to see with the rate at which the sum/series converges to its limit and the Roth-Thue-Siegel theorem which allows you to use the rate of convergence to decide if the limit is rational or not.



                                Maybe Emile (the OP) meant to ask something of this sort (please let me know, Emile): why do some (convergent, of course) infinite sums of rationals are rational and others are irrational?







                                share|cite|improve this answer














                                share|cite|improve this answer



                                share|cite|improve this answer








                                edited Mar 4 '12 at 10:25









                                g.castro

                                37218




                                37218










                                answered Mar 4 '12 at 10:09









                                AQP

                                12816




                                12816












                                • this is a different question..
                                  – AIB
                                  Mar 6 '12 at 9:41










                                • How so? The conditions/rate of convergence affect whether the limit is rational or not. Don't they?
                                  – AQP
                                  Mar 6 '12 at 16:54


















                                • this is a different question..
                                  – AIB
                                  Mar 6 '12 at 9:41










                                • How so? The conditions/rate of convergence affect whether the limit is rational or not. Don't they?
                                  – AQP
                                  Mar 6 '12 at 16:54
















                                this is a different question..
                                – AIB
                                Mar 6 '12 at 9:41




                                this is a different question..
                                – AIB
                                Mar 6 '12 at 9:41












                                How so? The conditions/rate of convergence affect whether the limit is rational or not. Don't they?
                                – AQP
                                Mar 6 '12 at 16:54




                                How so? The conditions/rate of convergence affect whether the limit is rational or not. Don't they?
                                – AQP
                                Mar 6 '12 at 16:54


















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