$sum_ n = 1^infty frac1(2n-1)(3n-1)$ The Next CEO of Stack OverflowSum of infinite series with arctan: $sum_n=1^inftyleft(arctanleft(frac14-nright)-arctanleft(-frac14-nright)right)$Prove the Series Converges or Diverges: $sum_n=1^infty frac5+2n(1+2n^2)^2$Why $sum_n=0^infty(-n)^3e^-n^-4$ absolutely converges?Convergence of the series $sum_n=1^inftyfrac12^n n (3n-1)$Hint on computing the series $sum_n=2^infty frac1n^2-1$.What does the series $sum_n=2^infty frac2n^3-n$ converge to?Compute the values of the Double SumHelp summing the telescoping series $sum_n=2^inftyfrac1n^3-n$.Evaluate $int_0^infty fract(t^2+1)(t^2+x^2) mathrmdt$if $f$ is such that : $int_-infty^infty x^2f(x) mathrmdx leq M$ then $f = O(x^3)$

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$sum_ n = 1^infty frac1(2n-1)(3n-1)$



The Next CEO of Stack OverflowSum of infinite series with arctan: $sum_n=1^inftyleft(arctanleft(frac14-nright)-arctanleft(-frac14-nright)right)$Prove the Series Converges or Diverges: $sum_n=1^infty frac5+2n(1+2n^2)^2$Why $sum_n=0^infty(-n)^3e^-n^-4$ absolutely converges?Convergence of the series $sum_n=1^inftyfrac12^n n (3n-1)$Hint on computing the series $sum_n=2^infty frac1n^2-1$.What does the series $sum_n=2^infty frac2n^3-n$ converge to?Compute the values of the Double SumHelp summing the telescoping series $sum_n=2^inftyfrac1n^3-n$.Evaluate $int_0^infty fract(t^2+1)(t^2+x^2) mathrmdt$if $f$ is such that : $int_-infty^infty x^2f(x) mathrmdx leq M$ then $f = O(x^3)$










1












$begingroup$


I would like to compute the value of the following sum




$$sum_ n = 1^infty frac1(2n-1)(3n-1)$$




Clearly, it converges since $ frac1(2n-1)(3n-1) = O(n^-2)$. I tried to use the partial fraction decomposition to get :
$$frac1(2n-1)(3n-1) = frac22n-1- frac33n-1$$



yet, it doesn't seem to lead anywhere since it's hard to see where the terms cancel out. So I don't really know what I could do in order to to attack this sum.



Thank you for your help!






real-analysis sequences-and-series limits







share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    Take a look at the properties of digamma function.
    $endgroup$
    – Math-fun
    Mar 19 at 11:36










  • $begingroup$
    If you just want the answer, Wolfram Alpha gives $$fracpi2sqrt 3+2log 2-frac32log 3$$ But I don't know how that was calculated.
    $endgroup$
    – TonyK
    Mar 19 at 11:37















1












$begingroup$


I would like to compute the value of the following sum




$$sum_ n = 1^infty frac1(2n-1)(3n-1)$$




Clearly, it converges since $ frac1(2n-1)(3n-1) = O(n^-2)$. I tried to use the partial fraction decomposition to get :
$$frac1(2n-1)(3n-1) = frac22n-1- frac33n-1$$



yet, it doesn't seem to lead anywhere since it's hard to see where the terms cancel out. So I don't really know what I could do in order to to attack this sum.



Thank you for your help!






real-analysis sequences-and-series limits







share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    Take a look at the properties of digamma function.
    $endgroup$
    – Math-fun
    Mar 19 at 11:36










  • $begingroup$
    If you just want the answer, Wolfram Alpha gives $$fracpi2sqrt 3+2log 2-frac32log 3$$ But I don't know how that was calculated.
    $endgroup$
    – TonyK
    Mar 19 at 11:37













1












1








1





$begingroup$


I would like to compute the value of the following sum




$$sum_ n = 1^infty frac1(2n-1)(3n-1)$$




Clearly, it converges since $ frac1(2n-1)(3n-1) = O(n^-2)$. I tried to use the partial fraction decomposition to get :
$$frac1(2n-1)(3n-1) = frac22n-1- frac33n-1$$



yet, it doesn't seem to lead anywhere since it's hard to see where the terms cancel out. So I don't really know what I could do in order to to attack this sum.



Thank you for your help!






real-analysis sequences-and-series limits







share|cite|improve this question











$endgroup$




I would like to compute the value of the following sum




$$sum_ n = 1^infty frac1(2n-1)(3n-1)$$




Clearly, it converges since $ frac1(2n-1)(3n-1) = O(n^-2)$. I tried to use the partial fraction decomposition to get :
$$frac1(2n-1)(3n-1) = frac22n-1- frac33n-1$$



yet, it doesn't seem to lead anywhere since it's hard to see where the terms cancel out. So I don't really know what I could do in order to to attack this sum.



Thank you for your help!






real-analysis sequences-and-series limits




real-analysis sequences-and-series limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 19 at 11:25









Chipa-Chipa

34




34










asked Mar 19 at 11:16









mouargmouargmouargmouarg

1145




1145







  • 1




    $begingroup$
    Take a look at the properties of digamma function.
    $endgroup$
    – Math-fun
    Mar 19 at 11:36










  • $begingroup$
    If you just want the answer, Wolfram Alpha gives $$fracpi2sqrt 3+2log 2-frac32log 3$$ But I don't know how that was calculated.
    $endgroup$
    – TonyK
    Mar 19 at 11:37












  • 1




    $begingroup$
    Take a look at the properties of digamma function.
    $endgroup$
    – Math-fun
    Mar 19 at 11:36










  • $begingroup$
    If you just want the answer, Wolfram Alpha gives $$fracpi2sqrt 3+2log 2-frac32log 3$$ But I don't know how that was calculated.
    $endgroup$
    – TonyK
    Mar 19 at 11:37







1




1




$begingroup$
Take a look at the properties of digamma function.
$endgroup$
– Math-fun
Mar 19 at 11:36




$begingroup$
Take a look at the properties of digamma function.
$endgroup$
– Math-fun
Mar 19 at 11:36












$begingroup$
If you just want the answer, Wolfram Alpha gives $$fracpi2sqrt 3+2log 2-frac32log 3$$ But I don't know how that was calculated.
$endgroup$
– TonyK
Mar 19 at 11:37




$begingroup$
If you just want the answer, Wolfram Alpha gives $$fracpi2sqrt 3+2log 2-frac32log 3$$ But I don't know how that was calculated.
$endgroup$
– TonyK
Mar 19 at 11:37










2 Answers
2






active

oldest

votes


















4












$begingroup$

As Math-fun commented, this is related to the digamma function
$$S_p=sum_ n = 1^p frac1(2n-1)(3n-1)=2sum_ n = 1^pfrac12n-1-3sum_ n = 1^p frac13n-1$$
$$S_p=psi ^(0)left(p+frac12right)-psi ^(0)left(p+frac23right)+psi
^(0)left(frac23right)-psi ^(0)left(frac12right)$$ Now, using the asymptotics
$$S_p=psi ^(0)left(frac23right)-psi
^(0)left(frac12right)-frac16 p+Oleft(frac1p^2right)$$ and
$$psi ^(0)left(frac23right)=-gamma +fracpi 2 sqrt3-frac3 log (3)2qquad textand qquad psi
^(0)left(frac12right)=-gamma -2log (2)$$






share|cite|improve this answer









$endgroup$




















    0












    $begingroup$

    Using they fact that
    $$psi(z+1)-psi(s+1)=sum_n=1^infty frac1n+s-frac1n+z$$
    Your sum is
    $$psi(2/3)-psi(1/2)$$
    This simplifies to
    $$fracpi2sqrt3-frac3log 32+2log 2$$






    share|cite|improve this answer









    $endgroup$













      Your Answer





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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      As Math-fun commented, this is related to the digamma function
      $$S_p=sum_ n = 1^p frac1(2n-1)(3n-1)=2sum_ n = 1^pfrac12n-1-3sum_ n = 1^p frac13n-1$$
      $$S_p=psi ^(0)left(p+frac12right)-psi ^(0)left(p+frac23right)+psi
      ^(0)left(frac23right)-psi ^(0)left(frac12right)$$       Now, using the asymptotics
      $$S_p=psi ^(0)left(frac23right)-psi
      ^(0)left(frac12right)-frac16 p+Oleft(frac1p^2right)$$       and
      $$psi ^(0)left(frac23right)=-gamma +fracpi 2 sqrt3-frac3 log (3)2qquad textand qquad psi
      ^(0)left(frac12right)=-gamma -2log (2)$$






      share|cite|improve this answer









      $endgroup$

















        4












        $begingroup$

        As Math-fun commented, this is related to the digamma function
        $$S_p=sum_ n = 1^p frac1(2n-1)(3n-1)=2sum_ n = 1^pfrac12n-1-3sum_ n = 1^p frac13n-1$$
        $$S_p=psi ^(0)left(p+frac12right)-psi ^(0)left(p+frac23right)+psi
        ^(0)left(frac23right)-psi ^(0)left(frac12right)$$         Now, using the asymptotics
        $$S_p=psi ^(0)left(frac23right)-psi
        ^(0)left(frac12right)-frac16 p+Oleft(frac1p^2right)$$         and
        $$psi ^(0)left(frac23right)=-gamma +fracpi 2 sqrt3-frac3 log (3)2qquad textand qquad psi
        ^(0)left(frac12right)=-gamma -2log (2)$$






        share|cite|improve this answer









        $endgroup$















          4












          4








          4





          $begingroup$

          As Math-fun commented, this is related to the digamma function
          $$S_p=sum_ n = 1^p frac1(2n-1)(3n-1)=2sum_ n = 1^pfrac12n-1-3sum_ n = 1^p frac13n-1$$
          $$S_p=psi ^(0)left(p+frac12right)-psi ^(0)left(p+frac23right)+psi
          ^(0)left(frac23right)-psi ^(0)left(frac12right)$$           Now, using the asymptotics
          $$S_p=psi ^(0)left(frac23right)-psi
          ^(0)left(frac12right)-frac16 p+Oleft(frac1p^2right)$$           and
          $$psi ^(0)left(frac23right)=-gamma +fracpi 2 sqrt3-frac3 log (3)2qquad textand qquad psi
          ^(0)left(frac12right)=-gamma -2log (2)$$






          share|cite|improve this answer









          $endgroup$



          As Math-fun commented, this is related to the digamma function
          $$S_p=sum_ n = 1^p frac1(2n-1)(3n-1)=2sum_ n = 1^pfrac12n-1-3sum_ n = 1^p frac13n-1$$
          $$S_p=psi ^(0)left(p+frac12right)-psi ^(0)left(p+frac23right)+psi
          ^(0)left(frac23right)-psi ^(0)left(frac12right)$$           Now, using the asymptotics
          $$S_p=psi ^(0)left(frac23right)-psi
          ^(0)left(frac12right)-frac16 p+Oleft(frac1p^2right)$$           and
          $$psi ^(0)left(frac23right)=-gamma +fracpi 2 sqrt3-frac3 log (3)2qquad textand qquad psi
          ^(0)left(frac12right)=-gamma -2log (2)$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 19 at 11:57









          Claude LeiboviciClaude Leibovici

          125k1158135




          125k1158135





















              0












              $begingroup$

              Using they fact that
              $$psi(z+1)-psi(s+1)=sum_n=1^infty frac1n+s-frac1n+z$$
              Your sum is
              $$psi(2/3)-psi(1/2)$$
              This simplifies to
              $$fracpi2sqrt3-frac3log 32+2log 2$$






              share|cite|improve this answer









              $endgroup$

















                0












                $begingroup$

                Using they fact that
                $$psi(z+1)-psi(s+1)=sum_n=1^infty frac1n+s-frac1n+z$$
                Your sum is
                $$psi(2/3)-psi(1/2)$$
                This simplifies to
                $$fracpi2sqrt3-frac3log 32+2log 2$$






                share|cite|improve this answer









                $endgroup$















                  0












                  0








                  0





                  $begingroup$

                  Using they fact that
                  $$psi(z+1)-psi(s+1)=sum_n=1^infty frac1n+s-frac1n+z$$
                  Your sum is
                  $$psi(2/3)-psi(1/2)$$
                  This simplifies to
                  $$fracpi2sqrt3-frac3log 32+2log 2$$






                  share|cite|improve this answer









                  $endgroup$



                  Using they fact that
                  $$psi(z+1)-psi(s+1)=sum_n=1^infty frac1n+s-frac1n+z$$
                  Your sum is
                  $$psi(2/3)-psi(1/2)$$
                  This simplifies to
                  $$fracpi2sqrt3-frac3log 32+2log 2$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 19 at 13:06









                  ZacharyZachary

                  2,3551214




                  2,3551214



























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