Prove that $ sumlimits_{i=1}^ infty frac{(-1)^{i+1}}{sqrt{i}}$ converges, using Abel's summation formula












0












$begingroup$


Let $(a_n)$ and $(b_n)$ be two sequences. Let $A_n = sum_{k=1}^n a_k$ and $A_0 = 0$. Prove that
$$sum_{n=p}^q a_n b_n = sum_{n=p}^{q-1} A_n(b_n - b_{n+1}) + A_qb_q - A_{p-1}b_p$$
for each $qgeq p geq 1$
Also, using this result prove that the series
$$ sum_{i=1}^ infty frac{(-1)^{i+1}}{sqrt{i}}$$
converges.



For the second, converge part, the series continues like $1,frac{-1}{sqrt2},frac{1}{sqrt3},frac{-1}{sqrt4}$. So it converges to 0. But I do not know how to prove first part also using the first part to show this series converges.






sequences-and-series summation-by-parts







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Start with $sum_{n=p}^q a_n b_n = sum_{n=p}^q A_n b_n - sum_{n=p}^q A_{n-1} b_n$. And $sum_{i=1}^ infty frac{(-1)^{i+1}}{sqrt{i}} ne 0$
    $endgroup$
    – reuns
    Nov 23 '18 at 5:10










  • $begingroup$
    Summation by parts
    $endgroup$
    – JavaMan
    Nov 23 '18 at 5:29
















0












$begingroup$


Let $(a_n)$ and $(b_n)$ be two sequences. Let $A_n = sum_{k=1}^n a_k$ and $A_0 = 0$. Prove that
$$sum_{n=p}^q a_n b_n = sum_{n=p}^{q-1} A_n(b_n - b_{n+1}) + A_qb_q - A_{p-1}b_p$$
for each $qgeq p geq 1$
Also, using this result prove that the series
$$ sum_{i=1}^ infty frac{(-1)^{i+1}}{sqrt{i}}$$
converges.



For the second, converge part, the series continues like $1,frac{-1}{sqrt2},frac{1}{sqrt3},frac{-1}{sqrt4}$. So it converges to 0. But I do not know how to prove first part also using the first part to show this series converges.






sequences-and-series summation-by-parts







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Start with $sum_{n=p}^q a_n b_n = sum_{n=p}^q A_n b_n - sum_{n=p}^q A_{n-1} b_n$. And $sum_{i=1}^ infty frac{(-1)^{i+1}}{sqrt{i}} ne 0$
    $endgroup$
    – reuns
    Nov 23 '18 at 5:10










  • $begingroup$
    Summation by parts
    $endgroup$
    – JavaMan
    Nov 23 '18 at 5:29














0












0








0


2



$begingroup$


Let $(a_n)$ and $(b_n)$ be two sequences. Let $A_n = sum_{k=1}^n a_k$ and $A_0 = 0$. Prove that
$$sum_{n=p}^q a_n b_n = sum_{n=p}^{q-1} A_n(b_n - b_{n+1}) + A_qb_q - A_{p-1}b_p$$
for each $qgeq p geq 1$
Also, using this result prove that the series
$$ sum_{i=1}^ infty frac{(-1)^{i+1}}{sqrt{i}}$$
converges.



For the second, converge part, the series continues like $1,frac{-1}{sqrt2},frac{1}{sqrt3},frac{-1}{sqrt4}$. So it converges to 0. But I do not know how to prove first part also using the first part to show this series converges.






sequences-and-series summation-by-parts







share|cite|improve this question











$endgroup$




Let $(a_n)$ and $(b_n)$ be two sequences. Let $A_n = sum_{k=1}^n a_k$ and $A_0 = 0$. Prove that
$$sum_{n=p}^q a_n b_n = sum_{n=p}^{q-1} A_n(b_n - b_{n+1}) + A_qb_q - A_{p-1}b_p$$
for each $qgeq p geq 1$
Also, using this result prove that the series
$$ sum_{i=1}^ infty frac{(-1)^{i+1}}{sqrt{i}}$$
converges.



For the second, converge part, the series continues like $1,frac{-1}{sqrt2},frac{1}{sqrt3},frac{-1}{sqrt4}$. So it converges to 0. But I do not know how to prove first part also using the first part to show this series converges.






sequences-and-series summation-by-parts




sequences-and-series summation-by-parts






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edited Dec 14 '18 at 11:28









Did

249k23228466




249k23228466










asked Nov 23 '18 at 5:03









PumpkinPumpkin

5021419




5021419








  • 1




    $begingroup$
    Start with $sum_{n=p}^q a_n b_n = sum_{n=p}^q A_n b_n - sum_{n=p}^q A_{n-1} b_n$. And $sum_{i=1}^ infty frac{(-1)^{i+1}}{sqrt{i}} ne 0$
    $endgroup$
    – reuns
    Nov 23 '18 at 5:10










  • $begingroup$
    Summation by parts
    $endgroup$
    – JavaMan
    Nov 23 '18 at 5:29














  • 1




    $begingroup$
    Start with $sum_{n=p}^q a_n b_n = sum_{n=p}^q A_n b_n - sum_{n=p}^q A_{n-1} b_n$. And $sum_{i=1}^ infty frac{(-1)^{i+1}}{sqrt{i}} ne 0$
    $endgroup$
    – reuns
    Nov 23 '18 at 5:10










  • $begingroup$
    Summation by parts
    $endgroup$
    – JavaMan
    Nov 23 '18 at 5:29








1




1




$begingroup$
Start with $sum_{n=p}^q a_n b_n = sum_{n=p}^q A_n b_n - sum_{n=p}^q A_{n-1} b_n$. And $sum_{i=1}^ infty frac{(-1)^{i+1}}{sqrt{i}} ne 0$
$endgroup$
– reuns
Nov 23 '18 at 5:10




$begingroup$
Start with $sum_{n=p}^q a_n b_n = sum_{n=p}^q A_n b_n - sum_{n=p}^q A_{n-1} b_n$. And $sum_{i=1}^ infty frac{(-1)^{i+1}}{sqrt{i}} ne 0$
$endgroup$
– reuns
Nov 23 '18 at 5:10












$begingroup$
Summation by parts
$endgroup$
– JavaMan
Nov 23 '18 at 5:29




$begingroup$
Summation by parts
$endgroup$
– JavaMan
Nov 23 '18 at 5:29










3 Answers
3






active

oldest

votes


















1












$begingroup$

Take $a_n=(-1)^{n+1}$ and $b_n=frac 1 {sqrt n}$. Check that $A_n$ is $1$ or $0$ for all $n$. Next note that $b_n-b_{n+1}=frac {sqrt {n+1}-sqrt {n}} {sqrt {n+1}sqrt {n}}$ which can be written as $frac 1 {sqrt {n+1}sqrt {n}(sqrt {n+1}+sqrt {n})} leq frac 1 {n^{3/2}}$. Use the fact that $sum frac 1 {n^{3/2}}<infty$ to complete the argument.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Answer for the second question:



    $sumlimits_{n=p}^q a_n b_n = sumlimits_{n=p}^{q-1} A_n(b_n - b_{n+1}) + A_qb_q - A_{p-1}b_p$



    Let's see the RHS of the statement:



    $ sumlimits_{n=p}^{q-1} A_nb_n - sumlimits_{n=p}^{q-1}A_nb_{n+1} + A_qb_q - A_{p-1}b_p$



    We can realize that



    $sumlimits_{n=p}^{q} A_nb_n=sumlimits_{n=p}^{q-1} A_nb_n+A_qb_q$ and



    $sumlimits_{n=p-1}^{q-1}A_nb_{n+1}=sumlimits_{n=p}^{q-1}A_nb_{n+1}+A_{p-1}b_p$



    So the RHS equal to:
    $sumlimits_{n=p}^{q} A_nb_n-sumlimits_{n=p-1}^{q-1}A_nb_{n+1}$



    After performing the reindex of the second term we get:
    $sumlimits_{n=p}^{q} A_nb_n-sumlimits_{n=p}^{q}A_{n-1}b_{n}$ and realize that



    $A_n - A_{n-1}=a_n$ we get the LHS of the statement: $sumlimits_{n=p}^q a_n b_n$






    share|cite|improve this answer









    $endgroup$





















      -1












      $begingroup$

      FYI, Determination of the value of the sum.



      $S=sumlimits_{i=1}^ infty frac{(-1)^{i+1}}{sqrt{i}}$



      Separate the sum accordance with even and odd terms. If i even then $i=2m$ and if i odd then $i=2m-1$



      $S=sumlimits_{m=1}^ inftybig( frac{1}{sqrt{2m-1}}-frac{1}{sqrt{2m}}big)=frac{1}{sqrt{2}}sumlimits_{m=0}^ inftybig( frac{1}{sqrt{m+frac{1}{2}}}-frac{1}{sqrt{m+1}}big)=frac{1}{sqrt{2}}sumlimits_{m=0}^ infty frac{1}{sqrt{m+frac{1}{2}}}-frac{1}{sqrt{2}}sumlimits_{m=1}^ inftyfrac{1}{sqrt{m}}$



      From the definition of the and Hurwitz zeta functions we get:



      $S=frac{1}{sqrt{2}}zeta(frac{1}{2},frac{1}{2})-frac{1}{sqrt{2}}zeta(frac{1}{2},1)$, as $zeta(frac{1}{2},1)=zeta(frac{1}{2})$



      and using the equality: $zeta(s,frac{1}{2})=(2^{-s}-1)zeta(s)$ (where $zeta(s)$ is the Riemann zeta function),
      we have the following result:



      $S=(1-sqrt{2})zeta(frac{1}{2})approx 0.60489...$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Unfortunately, both series $$sumlimits_{m=0}^ infty frac{1}{sqrt{m+frac{1}{2}}}$$ and $$sumlimits_{m=1}^ inftyfrac{1}{sqrt{m}}$$ diverge hence what this approach amounts to is to consider an indeterminate form $$infty-infty$$ as if it was not indeterminate. Not good!
        $endgroup$
        – Did
        Dec 14 '18 at 11:26














      Your Answer





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      3 Answers
      3






      active

      oldest

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      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Take $a_n=(-1)^{n+1}$ and $b_n=frac 1 {sqrt n}$. Check that $A_n$ is $1$ or $0$ for all $n$. Next note that $b_n-b_{n+1}=frac {sqrt {n+1}-sqrt {n}} {sqrt {n+1}sqrt {n}}$ which can be written as $frac 1 {sqrt {n+1}sqrt {n}(sqrt {n+1}+sqrt {n})} leq frac 1 {n^{3/2}}$. Use the fact that $sum frac 1 {n^{3/2}}<infty$ to complete the argument.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Take $a_n=(-1)^{n+1}$ and $b_n=frac 1 {sqrt n}$. Check that $A_n$ is $1$ or $0$ for all $n$. Next note that $b_n-b_{n+1}=frac {sqrt {n+1}-sqrt {n}} {sqrt {n+1}sqrt {n}}$ which can be written as $frac 1 {sqrt {n+1}sqrt {n}(sqrt {n+1}+sqrt {n})} leq frac 1 {n^{3/2}}$. Use the fact that $sum frac 1 {n^{3/2}}<infty$ to complete the argument.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Take $a_n=(-1)^{n+1}$ and $b_n=frac 1 {sqrt n}$. Check that $A_n$ is $1$ or $0$ for all $n$. Next note that $b_n-b_{n+1}=frac {sqrt {n+1}-sqrt {n}} {sqrt {n+1}sqrt {n}}$ which can be written as $frac 1 {sqrt {n+1}sqrt {n}(sqrt {n+1}+sqrt {n})} leq frac 1 {n^{3/2}}$. Use the fact that $sum frac 1 {n^{3/2}}<infty$ to complete the argument.






          share|cite|improve this answer









          $endgroup$



          Take $a_n=(-1)^{n+1}$ and $b_n=frac 1 {sqrt n}$. Check that $A_n$ is $1$ or $0$ for all $n$. Next note that $b_n-b_{n+1}=frac {sqrt {n+1}-sqrt {n}} {sqrt {n+1}sqrt {n}}$ which can be written as $frac 1 {sqrt {n+1}sqrt {n}(sqrt {n+1}+sqrt {n})} leq frac 1 {n^{3/2}}$. Use the fact that $sum frac 1 {n^{3/2}}<infty$ to complete the argument.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 23 '18 at 5:30









          Kavi Rama MurthyKavi Rama Murthy

          73.6k53170




          73.6k53170























              1












              $begingroup$

              Answer for the second question:



              $sumlimits_{n=p}^q a_n b_n = sumlimits_{n=p}^{q-1} A_n(b_n - b_{n+1}) + A_qb_q - A_{p-1}b_p$



              Let's see the RHS of the statement:



              $ sumlimits_{n=p}^{q-1} A_nb_n - sumlimits_{n=p}^{q-1}A_nb_{n+1} + A_qb_q - A_{p-1}b_p$



              We can realize that



              $sumlimits_{n=p}^{q} A_nb_n=sumlimits_{n=p}^{q-1} A_nb_n+A_qb_q$ and



              $sumlimits_{n=p-1}^{q-1}A_nb_{n+1}=sumlimits_{n=p}^{q-1}A_nb_{n+1}+A_{p-1}b_p$



              So the RHS equal to:
              $sumlimits_{n=p}^{q} A_nb_n-sumlimits_{n=p-1}^{q-1}A_nb_{n+1}$



              After performing the reindex of the second term we get:
              $sumlimits_{n=p}^{q} A_nb_n-sumlimits_{n=p}^{q}A_{n-1}b_{n}$ and realize that



              $A_n - A_{n-1}=a_n$ we get the LHS of the statement: $sumlimits_{n=p}^q a_n b_n$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Answer for the second question:



                $sumlimits_{n=p}^q a_n b_n = sumlimits_{n=p}^{q-1} A_n(b_n - b_{n+1}) + A_qb_q - A_{p-1}b_p$



                Let's see the RHS of the statement:



                $ sumlimits_{n=p}^{q-1} A_nb_n - sumlimits_{n=p}^{q-1}A_nb_{n+1} + A_qb_q - A_{p-1}b_p$



                We can realize that



                $sumlimits_{n=p}^{q} A_nb_n=sumlimits_{n=p}^{q-1} A_nb_n+A_qb_q$ and



                $sumlimits_{n=p-1}^{q-1}A_nb_{n+1}=sumlimits_{n=p}^{q-1}A_nb_{n+1}+A_{p-1}b_p$



                So the RHS equal to:
                $sumlimits_{n=p}^{q} A_nb_n-sumlimits_{n=p-1}^{q-1}A_nb_{n+1}$



                After performing the reindex of the second term we get:
                $sumlimits_{n=p}^{q} A_nb_n-sumlimits_{n=p}^{q}A_{n-1}b_{n}$ and realize that



                $A_n - A_{n-1}=a_n$ we get the LHS of the statement: $sumlimits_{n=p}^q a_n b_n$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Answer for the second question:



                  $sumlimits_{n=p}^q a_n b_n = sumlimits_{n=p}^{q-1} A_n(b_n - b_{n+1}) + A_qb_q - A_{p-1}b_p$



                  Let's see the RHS of the statement:



                  $ sumlimits_{n=p}^{q-1} A_nb_n - sumlimits_{n=p}^{q-1}A_nb_{n+1} + A_qb_q - A_{p-1}b_p$



                  We can realize that



                  $sumlimits_{n=p}^{q} A_nb_n=sumlimits_{n=p}^{q-1} A_nb_n+A_qb_q$ and



                  $sumlimits_{n=p-1}^{q-1}A_nb_{n+1}=sumlimits_{n=p}^{q-1}A_nb_{n+1}+A_{p-1}b_p$



                  So the RHS equal to:
                  $sumlimits_{n=p}^{q} A_nb_n-sumlimits_{n=p-1}^{q-1}A_nb_{n+1}$



                  After performing the reindex of the second term we get:
                  $sumlimits_{n=p}^{q} A_nb_n-sumlimits_{n=p}^{q}A_{n-1}b_{n}$ and realize that



                  $A_n - A_{n-1}=a_n$ we get the LHS of the statement: $sumlimits_{n=p}^q a_n b_n$






                  share|cite|improve this answer









                  $endgroup$



                  Answer for the second question:



                  $sumlimits_{n=p}^q a_n b_n = sumlimits_{n=p}^{q-1} A_n(b_n - b_{n+1}) + A_qb_q - A_{p-1}b_p$



                  Let's see the RHS of the statement:



                  $ sumlimits_{n=p}^{q-1} A_nb_n - sumlimits_{n=p}^{q-1}A_nb_{n+1} + A_qb_q - A_{p-1}b_p$



                  We can realize that



                  $sumlimits_{n=p}^{q} A_nb_n=sumlimits_{n=p}^{q-1} A_nb_n+A_qb_q$ and



                  $sumlimits_{n=p-1}^{q-1}A_nb_{n+1}=sumlimits_{n=p}^{q-1}A_nb_{n+1}+A_{p-1}b_p$



                  So the RHS equal to:
                  $sumlimits_{n=p}^{q} A_nb_n-sumlimits_{n=p-1}^{q-1}A_nb_{n+1}$



                  After performing the reindex of the second term we get:
                  $sumlimits_{n=p}^{q} A_nb_n-sumlimits_{n=p}^{q}A_{n-1}b_{n}$ and realize that



                  $A_n - A_{n-1}=a_n$ we get the LHS of the statement: $sumlimits_{n=p}^q a_n b_n$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 24 '18 at 7:01









                  JV.StalkerJV.Stalker

                  933149




                  933149























                      -1












                      $begingroup$

                      FYI, Determination of the value of the sum.



                      $S=sumlimits_{i=1}^ infty frac{(-1)^{i+1}}{sqrt{i}}$



                      Separate the sum accordance with even and odd terms. If i even then $i=2m$ and if i odd then $i=2m-1$



                      $S=sumlimits_{m=1}^ inftybig( frac{1}{sqrt{2m-1}}-frac{1}{sqrt{2m}}big)=frac{1}{sqrt{2}}sumlimits_{m=0}^ inftybig( frac{1}{sqrt{m+frac{1}{2}}}-frac{1}{sqrt{m+1}}big)=frac{1}{sqrt{2}}sumlimits_{m=0}^ infty frac{1}{sqrt{m+frac{1}{2}}}-frac{1}{sqrt{2}}sumlimits_{m=1}^ inftyfrac{1}{sqrt{m}}$



                      From the definition of the and Hurwitz zeta functions we get:



                      $S=frac{1}{sqrt{2}}zeta(frac{1}{2},frac{1}{2})-frac{1}{sqrt{2}}zeta(frac{1}{2},1)$, as $zeta(frac{1}{2},1)=zeta(frac{1}{2})$



                      and using the equality: $zeta(s,frac{1}{2})=(2^{-s}-1)zeta(s)$ (where $zeta(s)$ is the Riemann zeta function),
                      we have the following result:



                      $S=(1-sqrt{2})zeta(frac{1}{2})approx 0.60489...$






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        Unfortunately, both series $$sumlimits_{m=0}^ infty frac{1}{sqrt{m+frac{1}{2}}}$$ and $$sumlimits_{m=1}^ inftyfrac{1}{sqrt{m}}$$ diverge hence what this approach amounts to is to consider an indeterminate form $$infty-infty$$ as if it was not indeterminate. Not good!
                        $endgroup$
                        – Did
                        Dec 14 '18 at 11:26


















                      -1












                      $begingroup$

                      FYI, Determination of the value of the sum.



                      $S=sumlimits_{i=1}^ infty frac{(-1)^{i+1}}{sqrt{i}}$



                      Separate the sum accordance with even and odd terms. If i even then $i=2m$ and if i odd then $i=2m-1$



                      $S=sumlimits_{m=1}^ inftybig( frac{1}{sqrt{2m-1}}-frac{1}{sqrt{2m}}big)=frac{1}{sqrt{2}}sumlimits_{m=0}^ inftybig( frac{1}{sqrt{m+frac{1}{2}}}-frac{1}{sqrt{m+1}}big)=frac{1}{sqrt{2}}sumlimits_{m=0}^ infty frac{1}{sqrt{m+frac{1}{2}}}-frac{1}{sqrt{2}}sumlimits_{m=1}^ inftyfrac{1}{sqrt{m}}$



                      From the definition of the and Hurwitz zeta functions we get:



                      $S=frac{1}{sqrt{2}}zeta(frac{1}{2},frac{1}{2})-frac{1}{sqrt{2}}zeta(frac{1}{2},1)$, as $zeta(frac{1}{2},1)=zeta(frac{1}{2})$



                      and using the equality: $zeta(s,frac{1}{2})=(2^{-s}-1)zeta(s)$ (where $zeta(s)$ is the Riemann zeta function),
                      we have the following result:



                      $S=(1-sqrt{2})zeta(frac{1}{2})approx 0.60489...$






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        Unfortunately, both series $$sumlimits_{m=0}^ infty frac{1}{sqrt{m+frac{1}{2}}}$$ and $$sumlimits_{m=1}^ inftyfrac{1}{sqrt{m}}$$ diverge hence what this approach amounts to is to consider an indeterminate form $$infty-infty$$ as if it was not indeterminate. Not good!
                        $endgroup$
                        – Did
                        Dec 14 '18 at 11:26
















                      -1












                      -1








                      -1





                      $begingroup$

                      FYI, Determination of the value of the sum.



                      $S=sumlimits_{i=1}^ infty frac{(-1)^{i+1}}{sqrt{i}}$



                      Separate the sum accordance with even and odd terms. If i even then $i=2m$ and if i odd then $i=2m-1$



                      $S=sumlimits_{m=1}^ inftybig( frac{1}{sqrt{2m-1}}-frac{1}{sqrt{2m}}big)=frac{1}{sqrt{2}}sumlimits_{m=0}^ inftybig( frac{1}{sqrt{m+frac{1}{2}}}-frac{1}{sqrt{m+1}}big)=frac{1}{sqrt{2}}sumlimits_{m=0}^ infty frac{1}{sqrt{m+frac{1}{2}}}-frac{1}{sqrt{2}}sumlimits_{m=1}^ inftyfrac{1}{sqrt{m}}$



                      From the definition of the and Hurwitz zeta functions we get:



                      $S=frac{1}{sqrt{2}}zeta(frac{1}{2},frac{1}{2})-frac{1}{sqrt{2}}zeta(frac{1}{2},1)$, as $zeta(frac{1}{2},1)=zeta(frac{1}{2})$



                      and using the equality: $zeta(s,frac{1}{2})=(2^{-s}-1)zeta(s)$ (where $zeta(s)$ is the Riemann zeta function),
                      we have the following result:



                      $S=(1-sqrt{2})zeta(frac{1}{2})approx 0.60489...$






                      share|cite|improve this answer









                      $endgroup$



                      FYI, Determination of the value of the sum.



                      $S=sumlimits_{i=1}^ infty frac{(-1)^{i+1}}{sqrt{i}}$



                      Separate the sum accordance with even and odd terms. If i even then $i=2m$ and if i odd then $i=2m-1$



                      $S=sumlimits_{m=1}^ inftybig( frac{1}{sqrt{2m-1}}-frac{1}{sqrt{2m}}big)=frac{1}{sqrt{2}}sumlimits_{m=0}^ inftybig( frac{1}{sqrt{m+frac{1}{2}}}-frac{1}{sqrt{m+1}}big)=frac{1}{sqrt{2}}sumlimits_{m=0}^ infty frac{1}{sqrt{m+frac{1}{2}}}-frac{1}{sqrt{2}}sumlimits_{m=1}^ inftyfrac{1}{sqrt{m}}$



                      From the definition of the and Hurwitz zeta functions we get:



                      $S=frac{1}{sqrt{2}}zeta(frac{1}{2},frac{1}{2})-frac{1}{sqrt{2}}zeta(frac{1}{2},1)$, as $zeta(frac{1}{2},1)=zeta(frac{1}{2})$



                      and using the equality: $zeta(s,frac{1}{2})=(2^{-s}-1)zeta(s)$ (where $zeta(s)$ is the Riemann zeta function),
                      we have the following result:



                      $S=(1-sqrt{2})zeta(frac{1}{2})approx 0.60489...$







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                      answered Nov 23 '18 at 16:42









                      JV.StalkerJV.Stalker

                      933149




                      933149












                      • $begingroup$
                        Unfortunately, both series $$sumlimits_{m=0}^ infty frac{1}{sqrt{m+frac{1}{2}}}$$ and $$sumlimits_{m=1}^ inftyfrac{1}{sqrt{m}}$$ diverge hence what this approach amounts to is to consider an indeterminate form $$infty-infty$$ as if it was not indeterminate. Not good!
                        $endgroup$
                        – Did
                        Dec 14 '18 at 11:26




















                      • $begingroup$
                        Unfortunately, both series $$sumlimits_{m=0}^ infty frac{1}{sqrt{m+frac{1}{2}}}$$ and $$sumlimits_{m=1}^ inftyfrac{1}{sqrt{m}}$$ diverge hence what this approach amounts to is to consider an indeterminate form $$infty-infty$$ as if it was not indeterminate. Not good!
                        $endgroup$
                        – Did
                        Dec 14 '18 at 11:26


















                      $begingroup$
                      Unfortunately, both series $$sumlimits_{m=0}^ infty frac{1}{sqrt{m+frac{1}{2}}}$$ and $$sumlimits_{m=1}^ inftyfrac{1}{sqrt{m}}$$ diverge hence what this approach amounts to is to consider an indeterminate form $$infty-infty$$ as if it was not indeterminate. Not good!
                      $endgroup$
                      – Did
                      Dec 14 '18 at 11:26






                      $begingroup$
                      Unfortunately, both series $$sumlimits_{m=0}^ infty frac{1}{sqrt{m+frac{1}{2}}}$$ and $$sumlimits_{m=1}^ inftyfrac{1}{sqrt{m}}$$ diverge hence what this approach amounts to is to consider an indeterminate form $$infty-infty$$ as if it was not indeterminate. Not good!
                      $endgroup$
                      – Did
                      Dec 14 '18 at 11:26




















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