every real in (0,1) written as linear combination of the terms of a sequence












0












$begingroup$


Let $a_1, a_2, ..., a_n$ a infinite sequence of positive real numbers such that every real number in $(0,1)$ can be written as a linear combination with weights $n_1, n_2,...n_n in mathbb{N}$ of the $a_i$ ($a_1n_1+a_2n_2+a_3n_3+ ...$). How can I prove that $lim_{i rightarrow infty}a_i = 0$? And the viceversa is true?






real-analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It isn't true. Take any sequence ${b_n}$ that works and define a new sequence ${a_n}$ by $a_{2n}=b_n$ and $a_{2n+1}=1$. Then ${a_n}$ still has the property you want (since it contains the sequence ${b_n}$) but $a_n$ clearly does not approach $0$.
    $endgroup$
    – lulu
    Dec 1 '18 at 15:28












  • $begingroup$
    In my opinion the wording "linear combination" is misleading. It indicates that you only consider finite sums. But it seems that you allow any convergent series $sum_{k=1}^infty a_k n_k$.
    $endgroup$
    – Paul Frost
    Dec 1 '18 at 17:33










  • $begingroup$
    lulu's comment shows that the best you can expect is that there is a subsequence $(a_{i_k})$ such that $a_{i_k} to 0$ as $k to infty$.
    $endgroup$
    – Paul Frost
    Dec 1 '18 at 17:41
















0












$begingroup$


Let $a_1, a_2, ..., a_n$ a infinite sequence of positive real numbers such that every real number in $(0,1)$ can be written as a linear combination with weights $n_1, n_2,...n_n in mathbb{N}$ of the $a_i$ ($a_1n_1+a_2n_2+a_3n_3+ ...$). How can I prove that $lim_{i rightarrow infty}a_i = 0$? And the viceversa is true?






real-analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It isn't true. Take any sequence ${b_n}$ that works and define a new sequence ${a_n}$ by $a_{2n}=b_n$ and $a_{2n+1}=1$. Then ${a_n}$ still has the property you want (since it contains the sequence ${b_n}$) but $a_n$ clearly does not approach $0$.
    $endgroup$
    – lulu
    Dec 1 '18 at 15:28












  • $begingroup$
    In my opinion the wording "linear combination" is misleading. It indicates that you only consider finite sums. But it seems that you allow any convergent series $sum_{k=1}^infty a_k n_k$.
    $endgroup$
    – Paul Frost
    Dec 1 '18 at 17:33










  • $begingroup$
    lulu's comment shows that the best you can expect is that there is a subsequence $(a_{i_k})$ such that $a_{i_k} to 0$ as $k to infty$.
    $endgroup$
    – Paul Frost
    Dec 1 '18 at 17:41














0












0








0





$begingroup$


Let $a_1, a_2, ..., a_n$ a infinite sequence of positive real numbers such that every real number in $(0,1)$ can be written as a linear combination with weights $n_1, n_2,...n_n in mathbb{N}$ of the $a_i$ ($a_1n_1+a_2n_2+a_3n_3+ ...$). How can I prove that $lim_{i rightarrow infty}a_i = 0$? And the viceversa is true?






real-analysis







share|cite|improve this question











$endgroup$




Let $a_1, a_2, ..., a_n$ a infinite sequence of positive real numbers such that every real number in $(0,1)$ can be written as a linear combination with weights $n_1, n_2,...n_n in mathbb{N}$ of the $a_i$ ($a_1n_1+a_2n_2+a_3n_3+ ...$). How can I prove that $lim_{i rightarrow infty}a_i = 0$? And the viceversa is true?






real-analysis




real-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 1 '18 at 16:53







Lance

















asked Dec 1 '18 at 15:02









LanceLance

8912




8912








  • 1




    $begingroup$
    It isn't true. Take any sequence ${b_n}$ that works and define a new sequence ${a_n}$ by $a_{2n}=b_n$ and $a_{2n+1}=1$. Then ${a_n}$ still has the property you want (since it contains the sequence ${b_n}$) but $a_n$ clearly does not approach $0$.
    $endgroup$
    – lulu
    Dec 1 '18 at 15:28












  • $begingroup$
    In my opinion the wording "linear combination" is misleading. It indicates that you only consider finite sums. But it seems that you allow any convergent series $sum_{k=1}^infty a_k n_k$.
    $endgroup$
    – Paul Frost
    Dec 1 '18 at 17:33










  • $begingroup$
    lulu's comment shows that the best you can expect is that there is a subsequence $(a_{i_k})$ such that $a_{i_k} to 0$ as $k to infty$.
    $endgroup$
    – Paul Frost
    Dec 1 '18 at 17:41














  • 1




    $begingroup$
    It isn't true. Take any sequence ${b_n}$ that works and define a new sequence ${a_n}$ by $a_{2n}=b_n$ and $a_{2n+1}=1$. Then ${a_n}$ still has the property you want (since it contains the sequence ${b_n}$) but $a_n$ clearly does not approach $0$.
    $endgroup$
    – lulu
    Dec 1 '18 at 15:28












  • $begingroup$
    In my opinion the wording "linear combination" is misleading. It indicates that you only consider finite sums. But it seems that you allow any convergent series $sum_{k=1}^infty a_k n_k$.
    $endgroup$
    – Paul Frost
    Dec 1 '18 at 17:33










  • $begingroup$
    lulu's comment shows that the best you can expect is that there is a subsequence $(a_{i_k})$ such that $a_{i_k} to 0$ as $k to infty$.
    $endgroup$
    – Paul Frost
    Dec 1 '18 at 17:41








1




1




$begingroup$
It isn't true. Take any sequence ${b_n}$ that works and define a new sequence ${a_n}$ by $a_{2n}=b_n$ and $a_{2n+1}=1$. Then ${a_n}$ still has the property you want (since it contains the sequence ${b_n}$) but $a_n$ clearly does not approach $0$.
$endgroup$
– lulu
Dec 1 '18 at 15:28






$begingroup$
It isn't true. Take any sequence ${b_n}$ that works and define a new sequence ${a_n}$ by $a_{2n}=b_n$ and $a_{2n+1}=1$. Then ${a_n}$ still has the property you want (since it contains the sequence ${b_n}$) but $a_n$ clearly does not approach $0$.
$endgroup$
– lulu
Dec 1 '18 at 15:28














$begingroup$
In my opinion the wording "linear combination" is misleading. It indicates that you only consider finite sums. But it seems that you allow any convergent series $sum_{k=1}^infty a_k n_k$.
$endgroup$
– Paul Frost
Dec 1 '18 at 17:33




$begingroup$
In my opinion the wording "linear combination" is misleading. It indicates that you only consider finite sums. But it seems that you allow any convergent series $sum_{k=1}^infty a_k n_k$.
$endgroup$
– Paul Frost
Dec 1 '18 at 17:33












$begingroup$
lulu's comment shows that the best you can expect is that there is a subsequence $(a_{i_k})$ such that $a_{i_k} to 0$ as $k to infty$.
$endgroup$
– Paul Frost
Dec 1 '18 at 17:41




$begingroup$
lulu's comment shows that the best you can expect is that there is a subsequence $(a_{i_k})$ such that $a_{i_k} to 0$ as $k to infty$.
$endgroup$
– Paul Frost
Dec 1 '18 at 17:41










2 Answers
2






active

oldest

votes


















1












$begingroup$

If you regard $0$ as an element of $mathbb{N}$, then it is wrong ( see lulu's comment).



However, we can show that the following are equivalent:



(1) $(a_i)$ has a subsequence $(a_{i_k})$ such that $lim_{k to infty} a_{i_k} = 0$.



(2) Each $x in (0,1)$ can be represented as a convergent series $sum_{i=1}^infty a_i n_i$.



(1) $Rightarrow$ (2) : Let $x in (0,1)$. It suffices to consider the sequence $b_k = a_{i_k}$ (which converges to $0$) and construct $n_k$ such that $x = sum_{k=1}^infty b_k n_k$. This is done inductively.



$k = 1$ : If $x le b_1$, let $n_1 = 0$. If $b_1 < x$, let $n_1$ be the biggest number such that $b_1 n_1 < x$. In both cases $0 < x - b_1 n_1 le b_1$.



Assume we have constructed $n_1,dots,n_k$ such that $0 < x - sum_{r=1}^k b_r n_r le b_k$.



$k mapsto k+1$ : If $x - sum_{r=1}^k b_r n_r le b_{k+1}$, let $n_{k+1}= 0$. If $b_{k+1} < x - sum_{r=1}^k b_r n_r$, let $n_{k+1}$ be the biggest number such that $b_{k+1} n_{k+1} < x - sum_{r=1}^k b_r n_r$. In both cases $0 < x - sum_{r=1}^{k+1} b_r n_r le b_{k+1}$.



This shows that $x - sum_{r=1}^k b_r n_r to 0$ as $k to infty$. In other words, $x = sum_{k=1}^infty b_k n_k$.



(2) $Rightarrow$ (1) : There must exist $x in (0,1)$ such that in $x = sum_{i=1}^infty a_i n_i$ we have infinitely many $n_i ne 0$ (since the set of all $sum_{i=1}^infty a_i n_i$ in which only finitely many $n_i ne 0$ is countable). Let $n_{i_k}$ be the subsequence of nonzero weights. In a convergent series we have $a_i n_i to 0$, hence $a_{i_k} n_{i_k} to 0$ which implies $a_{i_k} to 0$.



Remark:



If you do not allow zero weights, then you do not have a chance to represent any $x in (0,1)$ as a convergent series. In fact, the minimal value would be $sum_{i=1}^infty a_i$.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    But if you assume that the weights are natural numbers it becomes true and follows from necessity condition of convergence of numbers series.
    By naturals I mean $1,2...$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      yes I forgot to add it. Anyway, the weights can be also 0
      $endgroup$
      – Lance
      Dec 1 '18 at 16:57











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    2 Answers
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    2 Answers
    2






    active

    oldest

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    active

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    active

    oldest

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    1












    $begingroup$

    If you regard $0$ as an element of $mathbb{N}$, then it is wrong ( see lulu's comment).



    However, we can show that the following are equivalent:



    (1) $(a_i)$ has a subsequence $(a_{i_k})$ such that $lim_{k to infty} a_{i_k} = 0$.



    (2) Each $x in (0,1)$ can be represented as a convergent series $sum_{i=1}^infty a_i n_i$.



    (1) $Rightarrow$ (2) : Let $x in (0,1)$. It suffices to consider the sequence $b_k = a_{i_k}$ (which converges to $0$) and construct $n_k$ such that $x = sum_{k=1}^infty b_k n_k$. This is done inductively.



    $k = 1$ : If $x le b_1$, let $n_1 = 0$. If $b_1 < x$, let $n_1$ be the biggest number such that $b_1 n_1 < x$. In both cases $0 < x - b_1 n_1 le b_1$.



    Assume we have constructed $n_1,dots,n_k$ such that $0 < x - sum_{r=1}^k b_r n_r le b_k$.



    $k mapsto k+1$ : If $x - sum_{r=1}^k b_r n_r le b_{k+1}$, let $n_{k+1}= 0$. If $b_{k+1} < x - sum_{r=1}^k b_r n_r$, let $n_{k+1}$ be the biggest number such that $b_{k+1} n_{k+1} < x - sum_{r=1}^k b_r n_r$. In both cases $0 < x - sum_{r=1}^{k+1} b_r n_r le b_{k+1}$.



    This shows that $x - sum_{r=1}^k b_r n_r to 0$ as $k to infty$. In other words, $x = sum_{k=1}^infty b_k n_k$.



    (2) $Rightarrow$ (1) : There must exist $x in (0,1)$ such that in $x = sum_{i=1}^infty a_i n_i$ we have infinitely many $n_i ne 0$ (since the set of all $sum_{i=1}^infty a_i n_i$ in which only finitely many $n_i ne 0$ is countable). Let $n_{i_k}$ be the subsequence of nonzero weights. In a convergent series we have $a_i n_i to 0$, hence $a_{i_k} n_{i_k} to 0$ which implies $a_{i_k} to 0$.



    Remark:



    If you do not allow zero weights, then you do not have a chance to represent any $x in (0,1)$ as a convergent series. In fact, the minimal value would be $sum_{i=1}^infty a_i$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      If you regard $0$ as an element of $mathbb{N}$, then it is wrong ( see lulu's comment).



      However, we can show that the following are equivalent:



      (1) $(a_i)$ has a subsequence $(a_{i_k})$ such that $lim_{k to infty} a_{i_k} = 0$.



      (2) Each $x in (0,1)$ can be represented as a convergent series $sum_{i=1}^infty a_i n_i$.



      (1) $Rightarrow$ (2) : Let $x in (0,1)$. It suffices to consider the sequence $b_k = a_{i_k}$ (which converges to $0$) and construct $n_k$ such that $x = sum_{k=1}^infty b_k n_k$. This is done inductively.



      $k = 1$ : If $x le b_1$, let $n_1 = 0$. If $b_1 < x$, let $n_1$ be the biggest number such that $b_1 n_1 < x$. In both cases $0 < x - b_1 n_1 le b_1$.



      Assume we have constructed $n_1,dots,n_k$ such that $0 < x - sum_{r=1}^k b_r n_r le b_k$.



      $k mapsto k+1$ : If $x - sum_{r=1}^k b_r n_r le b_{k+1}$, let $n_{k+1}= 0$. If $b_{k+1} < x - sum_{r=1}^k b_r n_r$, let $n_{k+1}$ be the biggest number such that $b_{k+1} n_{k+1} < x - sum_{r=1}^k b_r n_r$. In both cases $0 < x - sum_{r=1}^{k+1} b_r n_r le b_{k+1}$.



      This shows that $x - sum_{r=1}^k b_r n_r to 0$ as $k to infty$. In other words, $x = sum_{k=1}^infty b_k n_k$.



      (2) $Rightarrow$ (1) : There must exist $x in (0,1)$ such that in $x = sum_{i=1}^infty a_i n_i$ we have infinitely many $n_i ne 0$ (since the set of all $sum_{i=1}^infty a_i n_i$ in which only finitely many $n_i ne 0$ is countable). Let $n_{i_k}$ be the subsequence of nonzero weights. In a convergent series we have $a_i n_i to 0$, hence $a_{i_k} n_{i_k} to 0$ which implies $a_{i_k} to 0$.



      Remark:



      If you do not allow zero weights, then you do not have a chance to represent any $x in (0,1)$ as a convergent series. In fact, the minimal value would be $sum_{i=1}^infty a_i$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        If you regard $0$ as an element of $mathbb{N}$, then it is wrong ( see lulu's comment).



        However, we can show that the following are equivalent:



        (1) $(a_i)$ has a subsequence $(a_{i_k})$ such that $lim_{k to infty} a_{i_k} = 0$.



        (2) Each $x in (0,1)$ can be represented as a convergent series $sum_{i=1}^infty a_i n_i$.



        (1) $Rightarrow$ (2) : Let $x in (0,1)$. It suffices to consider the sequence $b_k = a_{i_k}$ (which converges to $0$) and construct $n_k$ such that $x = sum_{k=1}^infty b_k n_k$. This is done inductively.



        $k = 1$ : If $x le b_1$, let $n_1 = 0$. If $b_1 < x$, let $n_1$ be the biggest number such that $b_1 n_1 < x$. In both cases $0 < x - b_1 n_1 le b_1$.



        Assume we have constructed $n_1,dots,n_k$ such that $0 < x - sum_{r=1}^k b_r n_r le b_k$.



        $k mapsto k+1$ : If $x - sum_{r=1}^k b_r n_r le b_{k+1}$, let $n_{k+1}= 0$. If $b_{k+1} < x - sum_{r=1}^k b_r n_r$, let $n_{k+1}$ be the biggest number such that $b_{k+1} n_{k+1} < x - sum_{r=1}^k b_r n_r$. In both cases $0 < x - sum_{r=1}^{k+1} b_r n_r le b_{k+1}$.



        This shows that $x - sum_{r=1}^k b_r n_r to 0$ as $k to infty$. In other words, $x = sum_{k=1}^infty b_k n_k$.



        (2) $Rightarrow$ (1) : There must exist $x in (0,1)$ such that in $x = sum_{i=1}^infty a_i n_i$ we have infinitely many $n_i ne 0$ (since the set of all $sum_{i=1}^infty a_i n_i$ in which only finitely many $n_i ne 0$ is countable). Let $n_{i_k}$ be the subsequence of nonzero weights. In a convergent series we have $a_i n_i to 0$, hence $a_{i_k} n_{i_k} to 0$ which implies $a_{i_k} to 0$.



        Remark:



        If you do not allow zero weights, then you do not have a chance to represent any $x in (0,1)$ as a convergent series. In fact, the minimal value would be $sum_{i=1}^infty a_i$.






        share|cite|improve this answer











        $endgroup$



        If you regard $0$ as an element of $mathbb{N}$, then it is wrong ( see lulu's comment).



        However, we can show that the following are equivalent:



        (1) $(a_i)$ has a subsequence $(a_{i_k})$ such that $lim_{k to infty} a_{i_k} = 0$.



        (2) Each $x in (0,1)$ can be represented as a convergent series $sum_{i=1}^infty a_i n_i$.



        (1) $Rightarrow$ (2) : Let $x in (0,1)$. It suffices to consider the sequence $b_k = a_{i_k}$ (which converges to $0$) and construct $n_k$ such that $x = sum_{k=1}^infty b_k n_k$. This is done inductively.



        $k = 1$ : If $x le b_1$, let $n_1 = 0$. If $b_1 < x$, let $n_1$ be the biggest number such that $b_1 n_1 < x$. In both cases $0 < x - b_1 n_1 le b_1$.



        Assume we have constructed $n_1,dots,n_k$ such that $0 < x - sum_{r=1}^k b_r n_r le b_k$.



        $k mapsto k+1$ : If $x - sum_{r=1}^k b_r n_r le b_{k+1}$, let $n_{k+1}= 0$. If $b_{k+1} < x - sum_{r=1}^k b_r n_r$, let $n_{k+1}$ be the biggest number such that $b_{k+1} n_{k+1} < x - sum_{r=1}^k b_r n_r$. In both cases $0 < x - sum_{r=1}^{k+1} b_r n_r le b_{k+1}$.



        This shows that $x - sum_{r=1}^k b_r n_r to 0$ as $k to infty$. In other words, $x = sum_{k=1}^infty b_k n_k$.



        (2) $Rightarrow$ (1) : There must exist $x in (0,1)$ such that in $x = sum_{i=1}^infty a_i n_i$ we have infinitely many $n_i ne 0$ (since the set of all $sum_{i=1}^infty a_i n_i$ in which only finitely many $n_i ne 0$ is countable). Let $n_{i_k}$ be the subsequence of nonzero weights. In a convergent series we have $a_i n_i to 0$, hence $a_{i_k} n_{i_k} to 0$ which implies $a_{i_k} to 0$.



        Remark:



        If you do not allow zero weights, then you do not have a chance to represent any $x in (0,1)$ as a convergent series. In fact, the minimal value would be $sum_{i=1}^infty a_i$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 2 '18 at 9:27

























        answered Dec 1 '18 at 19:09









        Paul FrostPaul Frost

        9,9153932




        9,9153932























            0












            $begingroup$

            But if you assume that the weights are natural numbers it becomes true and follows from necessity condition of convergence of numbers series.
            By naturals I mean $1,2...$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              yes I forgot to add it. Anyway, the weights can be also 0
              $endgroup$
              – Lance
              Dec 1 '18 at 16:57
















            0












            $begingroup$

            But if you assume that the weights are natural numbers it becomes true and follows from necessity condition of convergence of numbers series.
            By naturals I mean $1,2...$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              yes I forgot to add it. Anyway, the weights can be also 0
              $endgroup$
              – Lance
              Dec 1 '18 at 16:57














            0












            0








            0





            $begingroup$

            But if you assume that the weights are natural numbers it becomes true and follows from necessity condition of convergence of numbers series.
            By naturals I mean $1,2...$






            share|cite|improve this answer









            $endgroup$



            But if you assume that the weights are natural numbers it becomes true and follows from necessity condition of convergence of numbers series.
            By naturals I mean $1,2...$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 1 '18 at 15:54









            MotylaNogaTomkaMazuraMotylaNogaTomkaMazura

            6,542917




            6,542917












            • $begingroup$
              yes I forgot to add it. Anyway, the weights can be also 0
              $endgroup$
              – Lance
              Dec 1 '18 at 16:57


















            • $begingroup$
              yes I forgot to add it. Anyway, the weights can be also 0
              $endgroup$
              – Lance
              Dec 1 '18 at 16:57
















            $begingroup$
            yes I forgot to add it. Anyway, the weights can be also 0
            $endgroup$
            – Lance
            Dec 1 '18 at 16:57




            $begingroup$
            yes I forgot to add it. Anyway, the weights can be also 0
            $endgroup$
            – Lance
            Dec 1 '18 at 16:57


















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