Finding $lim_{n to infty} 1+ 1/2 + 1/3 +dots +1/n - log n$. [duplicate]












2












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This question already has an answer here:




  • Simple proof Euler–Mascheroni $gamma$ constant

    2 answers




$$x_n = 1 + 1/2 +dots +1/n- log n$$



Then -



$1.$ Is the sequence increasing?



$2.$ is the sequence convergent?



For $(1)$, $sum 1/n$ is increasing and $log n $ is also increasing. First few terms are increasing, but i don't know about later terms.



$(2)$ $n^{th}$ term of the sequence can be written as $a_n = (sum_{i=1}^{n}) - log n$



So, $lim_{nto infty} a_n = lim_{n to infty} sum 1/n -lim_{n to infty} log n$



Neither first part nor second is convergent here. so i could not conclude anything.



How to solve?










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Dec 24 '18 at 17:54


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 4




    $begingroup$
    The sequence is found here: en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant
    $endgroup$
    – Zacky
    Dec 24 '18 at 14:49












  • $begingroup$
    Hint: consider that $log n = int_{1}^{n} frac{dx}{x}$. You can obtain the relevant estimates via Riemann sums.
    $endgroup$
    – MathematicsStudent1122
    Dec 24 '18 at 14:56












  • $begingroup$
    You have to be careful: $lim_n (a_n- b_n) = lim_n a_n - lim_n b_n$ is not true when $lim_n a_n =lim_n b_n=infty$.
    $endgroup$
    – Mircea
    Dec 24 '18 at 16:19










  • $begingroup$
    @Mircea , that's why I left the problem there.
    $endgroup$
    – Mathsaddict
    Dec 24 '18 at 16:29
















2












$begingroup$



This question already has an answer here:




  • Simple proof Euler–Mascheroni $gamma$ constant

    2 answers




$$x_n = 1 + 1/2 +dots +1/n- log n$$



Then -



$1.$ Is the sequence increasing?



$2.$ is the sequence convergent?



For $(1)$, $sum 1/n$ is increasing and $log n $ is also increasing. First few terms are increasing, but i don't know about later terms.



$(2)$ $n^{th}$ term of the sequence can be written as $a_n = (sum_{i=1}^{n}) - log n$



So, $lim_{nto infty} a_n = lim_{n to infty} sum 1/n -lim_{n to infty} log n$



Neither first part nor second is convergent here. so i could not conclude anything.



How to solve?










share|cite|improve this question











$endgroup$



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Dec 24 '18 at 17:54


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 4




    $begingroup$
    The sequence is found here: en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant
    $endgroup$
    – Zacky
    Dec 24 '18 at 14:49












  • $begingroup$
    Hint: consider that $log n = int_{1}^{n} frac{dx}{x}$. You can obtain the relevant estimates via Riemann sums.
    $endgroup$
    – MathematicsStudent1122
    Dec 24 '18 at 14:56












  • $begingroup$
    You have to be careful: $lim_n (a_n- b_n) = lim_n a_n - lim_n b_n$ is not true when $lim_n a_n =lim_n b_n=infty$.
    $endgroup$
    – Mircea
    Dec 24 '18 at 16:19










  • $begingroup$
    @Mircea , that's why I left the problem there.
    $endgroup$
    – Mathsaddict
    Dec 24 '18 at 16:29














2












2








2





$begingroup$



This question already has an answer here:




  • Simple proof Euler–Mascheroni $gamma$ constant

    2 answers




$$x_n = 1 + 1/2 +dots +1/n- log n$$



Then -



$1.$ Is the sequence increasing?



$2.$ is the sequence convergent?



For $(1)$, $sum 1/n$ is increasing and $log n $ is also increasing. First few terms are increasing, but i don't know about later terms.



$(2)$ $n^{th}$ term of the sequence can be written as $a_n = (sum_{i=1}^{n}) - log n$



So, $lim_{nto infty} a_n = lim_{n to infty} sum 1/n -lim_{n to infty} log n$



Neither first part nor second is convergent here. so i could not conclude anything.



How to solve?










share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Simple proof Euler–Mascheroni $gamma$ constant

    2 answers




$$x_n = 1 + 1/2 +dots +1/n- log n$$



Then -



$1.$ Is the sequence increasing?



$2.$ is the sequence convergent?



For $(1)$, $sum 1/n$ is increasing and $log n $ is also increasing. First few terms are increasing, but i don't know about later terms.



$(2)$ $n^{th}$ term of the sequence can be written as $a_n = (sum_{i=1}^{n}) - log n$



So, $lim_{nto infty} a_n = lim_{n to infty} sum 1/n -lim_{n to infty} log n$



Neither first part nor second is convergent here. so i could not conclude anything.



How to solve?





This question already has an answer here:




  • Simple proof Euler–Mascheroni $gamma$ constant

    2 answers








real-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 24 '18 at 15:51







Mathsaddict

















asked Dec 24 '18 at 14:46









MathsaddictMathsaddict

3669




3669




marked as duplicate by RRL real-analysis
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Dec 24 '18 at 17:54


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by RRL real-analysis
Users with the  real-analysis badge can single-handedly close real-analysis questions as duplicates and reopen them as needed.

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Dec 24 '18 at 17:54


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 4




    $begingroup$
    The sequence is found here: en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant
    $endgroup$
    – Zacky
    Dec 24 '18 at 14:49












  • $begingroup$
    Hint: consider that $log n = int_{1}^{n} frac{dx}{x}$. You can obtain the relevant estimates via Riemann sums.
    $endgroup$
    – MathematicsStudent1122
    Dec 24 '18 at 14:56












  • $begingroup$
    You have to be careful: $lim_n (a_n- b_n) = lim_n a_n - lim_n b_n$ is not true when $lim_n a_n =lim_n b_n=infty$.
    $endgroup$
    – Mircea
    Dec 24 '18 at 16:19










  • $begingroup$
    @Mircea , that's why I left the problem there.
    $endgroup$
    – Mathsaddict
    Dec 24 '18 at 16:29














  • 4




    $begingroup$
    The sequence is found here: en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant
    $endgroup$
    – Zacky
    Dec 24 '18 at 14:49












  • $begingroup$
    Hint: consider that $log n = int_{1}^{n} frac{dx}{x}$. You can obtain the relevant estimates via Riemann sums.
    $endgroup$
    – MathematicsStudent1122
    Dec 24 '18 at 14:56












  • $begingroup$
    You have to be careful: $lim_n (a_n- b_n) = lim_n a_n - lim_n b_n$ is not true when $lim_n a_n =lim_n b_n=infty$.
    $endgroup$
    – Mircea
    Dec 24 '18 at 16:19










  • $begingroup$
    @Mircea , that's why I left the problem there.
    $endgroup$
    – Mathsaddict
    Dec 24 '18 at 16:29








4




4




$begingroup$
The sequence is found here: en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant
$endgroup$
– Zacky
Dec 24 '18 at 14:49






$begingroup$
The sequence is found here: en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant
$endgroup$
– Zacky
Dec 24 '18 at 14:49














$begingroup$
Hint: consider that $log n = int_{1}^{n} frac{dx}{x}$. You can obtain the relevant estimates via Riemann sums.
$endgroup$
– MathematicsStudent1122
Dec 24 '18 at 14:56






$begingroup$
Hint: consider that $log n = int_{1}^{n} frac{dx}{x}$. You can obtain the relevant estimates via Riemann sums.
$endgroup$
– MathematicsStudent1122
Dec 24 '18 at 14:56














$begingroup$
You have to be careful: $lim_n (a_n- b_n) = lim_n a_n - lim_n b_n$ is not true when $lim_n a_n =lim_n b_n=infty$.
$endgroup$
– Mircea
Dec 24 '18 at 16:19




$begingroup$
You have to be careful: $lim_n (a_n- b_n) = lim_n a_n - lim_n b_n$ is not true when $lim_n a_n =lim_n b_n=infty$.
$endgroup$
– Mircea
Dec 24 '18 at 16:19












$begingroup$
@Mircea , that's why I left the problem there.
$endgroup$
– Mathsaddict
Dec 24 '18 at 16:29




$begingroup$
@Mircea , that's why I left the problem there.
$endgroup$
– Mathsaddict
Dec 24 '18 at 16:29










2 Answers
2






active

oldest

votes


















1












$begingroup$

This sequences converges to the Euler–Mascheroni constant.
It’s very important in number theory.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Let $x_n=-log(n)+sum_{k=1}^n frac1k$. Then, using $log(1+x)ge frac{x}{1+x}$, we see that



    $$begin{align}
    x_{n+1}-x_n&=frac1{n+1}-logleft(1+frac1nright)\\
    &le frac1{n+1}-frac{1}{n+1}\\
    &=0
    end{align}$$



    and $x_n$ is decreasing.





    Next, we can estimate the harmonic sum as $sum_{k=1}^n frac1kge frac12 sum_{k=1}^{n-1}left(frac1k+frac1{k+1}right)$, which represents the Trapezoidal Rule approximation of $int_1^n frac1x,dx$.



    Inasmuch as $frac1x$ is convex, the trapezoidal rule approximation overestimates the integral of $frac1x$ and we have



    $$sum_{k=1}^nfrac1k-log(n)ge sum_{k=1}^n frac1k -log(n)-frac12-frac1{2n}ge0 $$



    whence we see that



    $$x_nge frac12$$



    Since $x_n$ is decreasing and bounded below by $frac12$, the sequence $x_n$ converges.






    share|cite|improve this answer









    $endgroup$




















      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      This sequences converges to the Euler–Mascheroni constant.
      It’s very important in number theory.






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        This sequences converges to the Euler–Mascheroni constant.
        It’s very important in number theory.






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          This sequences converges to the Euler–Mascheroni constant.
          It’s very important in number theory.






          share|cite|improve this answer











          $endgroup$



          This sequences converges to the Euler–Mascheroni constant.
          It’s very important in number theory.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 24 '18 at 16:27









          Alex J Best

          2,44611227




          2,44611227










          answered Dec 24 '18 at 15:46









          Julio Trujillo GonzalezJulio Trujillo Gonzalez

          856




          856























              1












              $begingroup$

              Let $x_n=-log(n)+sum_{k=1}^n frac1k$. Then, using $log(1+x)ge frac{x}{1+x}$, we see that



              $$begin{align}
              x_{n+1}-x_n&=frac1{n+1}-logleft(1+frac1nright)\\
              &le frac1{n+1}-frac{1}{n+1}\\
              &=0
              end{align}$$



              and $x_n$ is decreasing.





              Next, we can estimate the harmonic sum as $sum_{k=1}^n frac1kge frac12 sum_{k=1}^{n-1}left(frac1k+frac1{k+1}right)$, which represents the Trapezoidal Rule approximation of $int_1^n frac1x,dx$.



              Inasmuch as $frac1x$ is convex, the trapezoidal rule approximation overestimates the integral of $frac1x$ and we have



              $$sum_{k=1}^nfrac1k-log(n)ge sum_{k=1}^n frac1k -log(n)-frac12-frac1{2n}ge0 $$



              whence we see that



              $$x_nge frac12$$



              Since $x_n$ is decreasing and bounded below by $frac12$, the sequence $x_n$ converges.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Let $x_n=-log(n)+sum_{k=1}^n frac1k$. Then, using $log(1+x)ge frac{x}{1+x}$, we see that



                $$begin{align}
                x_{n+1}-x_n&=frac1{n+1}-logleft(1+frac1nright)\\
                &le frac1{n+1}-frac{1}{n+1}\\
                &=0
                end{align}$$



                and $x_n$ is decreasing.





                Next, we can estimate the harmonic sum as $sum_{k=1}^n frac1kge frac12 sum_{k=1}^{n-1}left(frac1k+frac1{k+1}right)$, which represents the Trapezoidal Rule approximation of $int_1^n frac1x,dx$.



                Inasmuch as $frac1x$ is convex, the trapezoidal rule approximation overestimates the integral of $frac1x$ and we have



                $$sum_{k=1}^nfrac1k-log(n)ge sum_{k=1}^n frac1k -log(n)-frac12-frac1{2n}ge0 $$



                whence we see that



                $$x_nge frac12$$



                Since $x_n$ is decreasing and bounded below by $frac12$, the sequence $x_n$ converges.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Let $x_n=-log(n)+sum_{k=1}^n frac1k$. Then, using $log(1+x)ge frac{x}{1+x}$, we see that



                  $$begin{align}
                  x_{n+1}-x_n&=frac1{n+1}-logleft(1+frac1nright)\\
                  &le frac1{n+1}-frac{1}{n+1}\\
                  &=0
                  end{align}$$



                  and $x_n$ is decreasing.





                  Next, we can estimate the harmonic sum as $sum_{k=1}^n frac1kge frac12 sum_{k=1}^{n-1}left(frac1k+frac1{k+1}right)$, which represents the Trapezoidal Rule approximation of $int_1^n frac1x,dx$.



                  Inasmuch as $frac1x$ is convex, the trapezoidal rule approximation overestimates the integral of $frac1x$ and we have



                  $$sum_{k=1}^nfrac1k-log(n)ge sum_{k=1}^n frac1k -log(n)-frac12-frac1{2n}ge0 $$



                  whence we see that



                  $$x_nge frac12$$



                  Since $x_n$ is decreasing and bounded below by $frac12$, the sequence $x_n$ converges.






                  share|cite|improve this answer









                  $endgroup$



                  Let $x_n=-log(n)+sum_{k=1}^n frac1k$. Then, using $log(1+x)ge frac{x}{1+x}$, we see that



                  $$begin{align}
                  x_{n+1}-x_n&=frac1{n+1}-logleft(1+frac1nright)\\
                  &le frac1{n+1}-frac{1}{n+1}\\
                  &=0
                  end{align}$$



                  and $x_n$ is decreasing.





                  Next, we can estimate the harmonic sum as $sum_{k=1}^n frac1kge frac12 sum_{k=1}^{n-1}left(frac1k+frac1{k+1}right)$, which represents the Trapezoidal Rule approximation of $int_1^n frac1x,dx$.



                  Inasmuch as $frac1x$ is convex, the trapezoidal rule approximation overestimates the integral of $frac1x$ and we have



                  $$sum_{k=1}^nfrac1k-log(n)ge sum_{k=1}^n frac1k -log(n)-frac12-frac1{2n}ge0 $$



                  whence we see that



                  $$x_nge frac12$$



                  Since $x_n$ is decreasing and bounded below by $frac12$, the sequence $x_n$ converges.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 24 '18 at 17:42









                  Mark ViolaMark Viola

                  134k1278177




                  134k1278177















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