Evaluate $sum_{n=2}^infty frac{ln(n)(-1)^n}{n^2} $












3














I'm trying to find a closed for for the sum:
$$sum_{n=2}^infty frac{ln(n)(-1)^n}{n^2} $$



It's been proven, $$sum_{n=2}^infty frac{ln(n)(-1)^n}{n} = gammaln(2)-frac{ln(2)^2}{2}$$



Since they're similar, I feel there potentially could be a closed form solution of the first one. I want to know if there is also a closed form solution for the first summation.



I managed to rewrite the first sum as a double integral:



$$-int_0^1int_0^1 frac{ln(x)}{ln(y)}left(frac{1}{1+x}-frac{1}{1+xy} right)dxdy $$
But I don't see any overt way of evaluating it. I've tried differentiating under the integral, but to no avail. I can't think of any change of variables that would help simplify as well.






calculus integration sequences-and-series summation







share|cite|improve this question


















  • 4




    Well, $sum_{n=2}^infty frac{ln(n)(-1)^n}{n} $ is the derivative of eta funtion in $1$, see here: en.wikipedia.org/wiki/Dirichlet_eta_function#Derivatives. Why not plug $s=2$?
    – Zacky
    Jan 4 at 15:22












  • I see thank you
    – Tom Himler
    Jan 4 at 15:26






  • 2




    You're welcome, but there is still some work to do. From here:en.wikipedia.org/wiki/Glaisher%E2%80%93Kinkelin_constant we can find a value for $zeta'(2)$ and we should arrive at: $$sum_{n=2}^infty frac{ln(n)(-1)^n}{n}=eta'(2)=frac{pi^2}{12}left(ln(4pi) -12ln(A) +gammaright)$$
    – Zacky
    Jan 4 at 15:34








  • 1




    Just so you know, in your sum I think you forgot to square the denominator. Right now, the sum reads $eta'(1)$.
    – Zachary
    Jan 4 at 16:41










  • From differentiating $eta(s)=(1-2^{1-s})zeta(s)$, we obtain $eta'(s)=zeta(s)2^{1-s}log 2+big(1-2^{1-s}big)zeta'(s)$. However, I am not sure what happens near $s=1$, as the left side should be well defined, but the right side does not seem to be.
    – Zachary
    Jan 4 at 22:35


















3














I'm trying to find a closed for for the sum:
$$sum_{n=2}^infty frac{ln(n)(-1)^n}{n^2} $$



It's been proven, $$sum_{n=2}^infty frac{ln(n)(-1)^n}{n} = gammaln(2)-frac{ln(2)^2}{2}$$



Since they're similar, I feel there potentially could be a closed form solution of the first one. I want to know if there is also a closed form solution for the first summation.



I managed to rewrite the first sum as a double integral:



$$-int_0^1int_0^1 frac{ln(x)}{ln(y)}left(frac{1}{1+x}-frac{1}{1+xy} right)dxdy $$
But I don't see any overt way of evaluating it. I've tried differentiating under the integral, but to no avail. I can't think of any change of variables that would help simplify as well.






calculus integration sequences-and-series summation







share|cite|improve this question


















  • 4




    Well, $sum_{n=2}^infty frac{ln(n)(-1)^n}{n} $ is the derivative of eta funtion in $1$, see here: en.wikipedia.org/wiki/Dirichlet_eta_function#Derivatives. Why not plug $s=2$?
    – Zacky
    Jan 4 at 15:22












  • I see thank you
    – Tom Himler
    Jan 4 at 15:26






  • 2




    You're welcome, but there is still some work to do. From here:en.wikipedia.org/wiki/Glaisher%E2%80%93Kinkelin_constant we can find a value for $zeta'(2)$ and we should arrive at: $$sum_{n=2}^infty frac{ln(n)(-1)^n}{n}=eta'(2)=frac{pi^2}{12}left(ln(4pi) -12ln(A) +gammaright)$$
    – Zacky
    Jan 4 at 15:34








  • 1




    Just so you know, in your sum I think you forgot to square the denominator. Right now, the sum reads $eta'(1)$.
    – Zachary
    Jan 4 at 16:41










  • From differentiating $eta(s)=(1-2^{1-s})zeta(s)$, we obtain $eta'(s)=zeta(s)2^{1-s}log 2+big(1-2^{1-s}big)zeta'(s)$. However, I am not sure what happens near $s=1$, as the left side should be well defined, but the right side does not seem to be.
    – Zachary
    Jan 4 at 22:35
















3












3








3


1





I'm trying to find a closed for for the sum:
$$sum_{n=2}^infty frac{ln(n)(-1)^n}{n^2} $$



It's been proven, $$sum_{n=2}^infty frac{ln(n)(-1)^n}{n} = gammaln(2)-frac{ln(2)^2}{2}$$



Since they're similar, I feel there potentially could be a closed form solution of the first one. I want to know if there is also a closed form solution for the first summation.



I managed to rewrite the first sum as a double integral:



$$-int_0^1int_0^1 frac{ln(x)}{ln(y)}left(frac{1}{1+x}-frac{1}{1+xy} right)dxdy $$
But I don't see any overt way of evaluating it. I've tried differentiating under the integral, but to no avail. I can't think of any change of variables that would help simplify as well.






calculus integration sequences-and-series summation







share|cite|improve this question













I'm trying to find a closed for for the sum:
$$sum_{n=2}^infty frac{ln(n)(-1)^n}{n^2} $$



It's been proven, $$sum_{n=2}^infty frac{ln(n)(-1)^n}{n} = gammaln(2)-frac{ln(2)^2}{2}$$



Since they're similar, I feel there potentially could be a closed form solution of the first one. I want to know if there is also a closed form solution for the first summation.



I managed to rewrite the first sum as a double integral:



$$-int_0^1int_0^1 frac{ln(x)}{ln(y)}left(frac{1}{1+x}-frac{1}{1+xy} right)dxdy $$
But I don't see any overt way of evaluating it. I've tried differentiating under the integral, but to no avail. I can't think of any change of variables that would help simplify as well.






calculus integration sequences-and-series summation




calculus integration sequences-and-series summation






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 4 at 15:18









Tom HimlerTom Himler

902213




902213








  • 4




    Well, $sum_{n=2}^infty frac{ln(n)(-1)^n}{n} $ is the derivative of eta funtion in $1$, see here: en.wikipedia.org/wiki/Dirichlet_eta_function#Derivatives. Why not plug $s=2$?
    – Zacky
    Jan 4 at 15:22












  • I see thank you
    – Tom Himler
    Jan 4 at 15:26






  • 2




    You're welcome, but there is still some work to do. From here:en.wikipedia.org/wiki/Glaisher%E2%80%93Kinkelin_constant we can find a value for $zeta'(2)$ and we should arrive at: $$sum_{n=2}^infty frac{ln(n)(-1)^n}{n}=eta'(2)=frac{pi^2}{12}left(ln(4pi) -12ln(A) +gammaright)$$
    – Zacky
    Jan 4 at 15:34








  • 1




    Just so you know, in your sum I think you forgot to square the denominator. Right now, the sum reads $eta'(1)$.
    – Zachary
    Jan 4 at 16:41










  • From differentiating $eta(s)=(1-2^{1-s})zeta(s)$, we obtain $eta'(s)=zeta(s)2^{1-s}log 2+big(1-2^{1-s}big)zeta'(s)$. However, I am not sure what happens near $s=1$, as the left side should be well defined, but the right side does not seem to be.
    – Zachary
    Jan 4 at 22:35
















  • 4




    Well, $sum_{n=2}^infty frac{ln(n)(-1)^n}{n} $ is the derivative of eta funtion in $1$, see here: en.wikipedia.org/wiki/Dirichlet_eta_function#Derivatives. Why not plug $s=2$?
    – Zacky
    Jan 4 at 15:22












  • I see thank you
    – Tom Himler
    Jan 4 at 15:26






  • 2




    You're welcome, but there is still some work to do. From here:en.wikipedia.org/wiki/Glaisher%E2%80%93Kinkelin_constant we can find a value for $zeta'(2)$ and we should arrive at: $$sum_{n=2}^infty frac{ln(n)(-1)^n}{n}=eta'(2)=frac{pi^2}{12}left(ln(4pi) -12ln(A) +gammaright)$$
    – Zacky
    Jan 4 at 15:34








  • 1




    Just so you know, in your sum I think you forgot to square the denominator. Right now, the sum reads $eta'(1)$.
    – Zachary
    Jan 4 at 16:41










  • From differentiating $eta(s)=(1-2^{1-s})zeta(s)$, we obtain $eta'(s)=zeta(s)2^{1-s}log 2+big(1-2^{1-s}big)zeta'(s)$. However, I am not sure what happens near $s=1$, as the left side should be well defined, but the right side does not seem to be.
    – Zachary
    Jan 4 at 22:35










4




4




Well, $sum_{n=2}^infty frac{ln(n)(-1)^n}{n} $ is the derivative of eta funtion in $1$, see here: en.wikipedia.org/wiki/Dirichlet_eta_function#Derivatives. Why not plug $s=2$?
– Zacky
Jan 4 at 15:22






Well, $sum_{n=2}^infty frac{ln(n)(-1)^n}{n} $ is the derivative of eta funtion in $1$, see here: en.wikipedia.org/wiki/Dirichlet_eta_function#Derivatives. Why not plug $s=2$?
– Zacky
Jan 4 at 15:22














I see thank you
– Tom Himler
Jan 4 at 15:26




I see thank you
– Tom Himler
Jan 4 at 15:26




2




2




You're welcome, but there is still some work to do. From here:en.wikipedia.org/wiki/Glaisher%E2%80%93Kinkelin_constant we can find a value for $zeta'(2)$ and we should arrive at: $$sum_{n=2}^infty frac{ln(n)(-1)^n}{n}=eta'(2)=frac{pi^2}{12}left(ln(4pi) -12ln(A) +gammaright)$$
– Zacky
Jan 4 at 15:34






You're welcome, but there is still some work to do. From here:en.wikipedia.org/wiki/Glaisher%E2%80%93Kinkelin_constant we can find a value for $zeta'(2)$ and we should arrive at: $$sum_{n=2}^infty frac{ln(n)(-1)^n}{n}=eta'(2)=frac{pi^2}{12}left(ln(4pi) -12ln(A) +gammaright)$$
– Zacky
Jan 4 at 15:34






1




1




Just so you know, in your sum I think you forgot to square the denominator. Right now, the sum reads $eta'(1)$.
– Zachary
Jan 4 at 16:41




Just so you know, in your sum I think you forgot to square the denominator. Right now, the sum reads $eta'(1)$.
– Zachary
Jan 4 at 16:41












From differentiating $eta(s)=(1-2^{1-s})zeta(s)$, we obtain $eta'(s)=zeta(s)2^{1-s}log 2+big(1-2^{1-s}big)zeta'(s)$. However, I am not sure what happens near $s=1$, as the left side should be well defined, but the right side does not seem to be.
– Zachary
Jan 4 at 22:35






From differentiating $eta(s)=(1-2^{1-s})zeta(s)$, we obtain $eta'(s)=zeta(s)2^{1-s}log 2+big(1-2^{1-s}big)zeta'(s)$. However, I am not sure what happens near $s=1$, as the left side should be well defined, but the right side does not seem to be.
– Zachary
Jan 4 at 22:35












1 Answer
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The definition of $eta(s)$ is
$$eta(s)=(1-2^{1-s})zeta(s)$$
Take the derivative, and we get by product rule
$$eta'(s)=2^{1-s}ln(2)zeta(s)+(1-2^{1-s})zeta'(s)$$
$eta(s)$ also has the following series representation
$$eta(s)=sum_{n=1}^{infty}frac{(-1)^{n-1}}{n^s}$$
Again, take the derivative
$$eta'(s)=sum_{n=2}^{infty}frac{(-1)^{n}ln(n)}{n^s}$$
Now, we can obtain a close form
$$eta'(2)=sum_{n=2}^{infty}frac{(-1)^{n}ln(n)}{n^2}=2^{1-2}ln(2)zeta(2)+(1-2^{1-2})zeta'(2)$$
Note that




$$zeta'(2)=frac{1}{6}pi^2(-12ln(A)+ln(2pi)+gamma),~zeta(2)=frac{pi^2}{6}$$




$$begin{align}
sum_{n=2}^{infty}frac{(-1)^{n}ln(n)}{n^2}&=2^{-1}ln(2)zeta(2)+(1-2^{-1})zeta'(2)\
&=frac{pi^2}{12}ln(2)+frac{pi^2}{12}(-12ln(A)+ln(2pi)+gamma)\
&=frac{pi^2}{12}(ln(4pi)-12ln(A)+gamma)
end{align}$$






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    6














    The definition of $eta(s)$ is
    $$eta(s)=(1-2^{1-s})zeta(s)$$
    Take the derivative, and we get by product rule
    $$eta'(s)=2^{1-s}ln(2)zeta(s)+(1-2^{1-s})zeta'(s)$$
    $eta(s)$ also has the following series representation
    $$eta(s)=sum_{n=1}^{infty}frac{(-1)^{n-1}}{n^s}$$
    Again, take the derivative
    $$eta'(s)=sum_{n=2}^{infty}frac{(-1)^{n}ln(n)}{n^s}$$
    Now, we can obtain a close form
    $$eta'(2)=sum_{n=2}^{infty}frac{(-1)^{n}ln(n)}{n^2}=2^{1-2}ln(2)zeta(2)+(1-2^{1-2})zeta'(2)$$
    Note that




    $$zeta'(2)=frac{1}{6}pi^2(-12ln(A)+ln(2pi)+gamma),~zeta(2)=frac{pi^2}{6}$$




    $$begin{align}
    sum_{n=2}^{infty}frac{(-1)^{n}ln(n)}{n^2}&=2^{-1}ln(2)zeta(2)+(1-2^{-1})zeta'(2)\
    &=frac{pi^2}{12}ln(2)+frac{pi^2}{12}(-12ln(A)+ln(2pi)+gamma)\
    &=frac{pi^2}{12}(ln(4pi)-12ln(A)+gamma)
    end{align}$$






    share|cite|improve this answer


























      6














      The definition of $eta(s)$ is
      $$eta(s)=(1-2^{1-s})zeta(s)$$
      Take the derivative, and we get by product rule
      $$eta'(s)=2^{1-s}ln(2)zeta(s)+(1-2^{1-s})zeta'(s)$$
      $eta(s)$ also has the following series representation
      $$eta(s)=sum_{n=1}^{infty}frac{(-1)^{n-1}}{n^s}$$
      Again, take the derivative
      $$eta'(s)=sum_{n=2}^{infty}frac{(-1)^{n}ln(n)}{n^s}$$
      Now, we can obtain a close form
      $$eta'(2)=sum_{n=2}^{infty}frac{(-1)^{n}ln(n)}{n^2}=2^{1-2}ln(2)zeta(2)+(1-2^{1-2})zeta'(2)$$
      Note that




      $$zeta'(2)=frac{1}{6}pi^2(-12ln(A)+ln(2pi)+gamma),~zeta(2)=frac{pi^2}{6}$$




      $$begin{align}
      sum_{n=2}^{infty}frac{(-1)^{n}ln(n)}{n^2}&=2^{-1}ln(2)zeta(2)+(1-2^{-1})zeta'(2)\
      &=frac{pi^2}{12}ln(2)+frac{pi^2}{12}(-12ln(A)+ln(2pi)+gamma)\
      &=frac{pi^2}{12}(ln(4pi)-12ln(A)+gamma)
      end{align}$$






      share|cite|improve this answer
























        6












        6








        6






        The definition of $eta(s)$ is
        $$eta(s)=(1-2^{1-s})zeta(s)$$
        Take the derivative, and we get by product rule
        $$eta'(s)=2^{1-s}ln(2)zeta(s)+(1-2^{1-s})zeta'(s)$$
        $eta(s)$ also has the following series representation
        $$eta(s)=sum_{n=1}^{infty}frac{(-1)^{n-1}}{n^s}$$
        Again, take the derivative
        $$eta'(s)=sum_{n=2}^{infty}frac{(-1)^{n}ln(n)}{n^s}$$
        Now, we can obtain a close form
        $$eta'(2)=sum_{n=2}^{infty}frac{(-1)^{n}ln(n)}{n^2}=2^{1-2}ln(2)zeta(2)+(1-2^{1-2})zeta'(2)$$
        Note that




        $$zeta'(2)=frac{1}{6}pi^2(-12ln(A)+ln(2pi)+gamma),~zeta(2)=frac{pi^2}{6}$$




        $$begin{align}
        sum_{n=2}^{infty}frac{(-1)^{n}ln(n)}{n^2}&=2^{-1}ln(2)zeta(2)+(1-2^{-1})zeta'(2)\
        &=frac{pi^2}{12}ln(2)+frac{pi^2}{12}(-12ln(A)+ln(2pi)+gamma)\
        &=frac{pi^2}{12}(ln(4pi)-12ln(A)+gamma)
        end{align}$$






        share|cite|improve this answer












        The definition of $eta(s)$ is
        $$eta(s)=(1-2^{1-s})zeta(s)$$
        Take the derivative, and we get by product rule
        $$eta'(s)=2^{1-s}ln(2)zeta(s)+(1-2^{1-s})zeta'(s)$$
        $eta(s)$ also has the following series representation
        $$eta(s)=sum_{n=1}^{infty}frac{(-1)^{n-1}}{n^s}$$
        Again, take the derivative
        $$eta'(s)=sum_{n=2}^{infty}frac{(-1)^{n}ln(n)}{n^s}$$
        Now, we can obtain a close form
        $$eta'(2)=sum_{n=2}^{infty}frac{(-1)^{n}ln(n)}{n^2}=2^{1-2}ln(2)zeta(2)+(1-2^{1-2})zeta'(2)$$
        Note that




        $$zeta'(2)=frac{1}{6}pi^2(-12ln(A)+ln(2pi)+gamma),~zeta(2)=frac{pi^2}{6}$$




        $$begin{align}
        sum_{n=2}^{infty}frac{(-1)^{n}ln(n)}{n^2}&=2^{-1}ln(2)zeta(2)+(1-2^{-1})zeta'(2)\
        &=frac{pi^2}{12}ln(2)+frac{pi^2}{12}(-12ln(A)+ln(2pi)+gamma)\
        &=frac{pi^2}{12}(ln(4pi)-12ln(A)+gamma)
        end{align}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 4 at 23:46









        LarryLarry

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