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How to find the sum of

$$\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\left\{\sum_{k=1}^{2n+1}\frac{(-1)^k} k \right\}$$

$$\begin{array}\\ \frac{1}{1}&\times&(-\frac{1}{1}) &+&\ \ \\ (-\frac{1}{3})&\times&(-\frac{1}{1}&+&\frac{1}{2}&-&\frac{1}{3})&+&\\ \frac{1}{5}&\times&(-\frac{1}{1}&+&\frac{1}{2}&-&\frac{1}{3}&+&\frac{1}{4}&-&\frac{1}{5})&+&\\ (-\frac{1}{7})&\times&(-\frac{1}{1}&+&\frac{1}{2}&-&\frac{1}{3}&+&\frac{1}{4}&-&\frac{1}{5}&+&\frac{1}{6}&-&\frac{1}{7})&+&\\ \\ & \,\vdots \end{array}$$

I have no idea how to start this question.(Virtually,I have a basic idea of related theories about double series. )Effective hints should be necessary for me.

james givens
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2 Answers2

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The inner sum can be written as $$-\int_{0}^{1}\sum_{k=0}^{2n}(-1)^{k}x^{k}\,dx$$ which simplifies to $$-\int_{0}^{1}\frac{1+x^{2n+1}}{1+x}\,dx$$ or $$-\log 2-\int_{0}^{1}\frac{x^{2n+1}}{1+x}\,dx$$ Thus the sum in question is equal to $$-\frac{\pi\log 2}{4}-\int_{0}^{1}\frac{1}{1+x}\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{2n+1}\,dx$$ (note the interchange of summation and integration which can be justified using very general theorems if needed) which is same as $$-\frac{\pi\log 2}{4}-\int_{0}^{1}\frac{\arctan x} {1+x}\,dx$$ I hope the integral can be evaluated in closed form and you should try to proceed from here. Based on comment from OP, we can see that the sum of the series is $$-\frac{3\pi\log 2}{8}$$ and this also matches the numerical value obtained in another answer from user "Claude Leibovici".

Paramanand Singh
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    The closed from will be get from your efforts.Because $$ \int_{0}^{1}\frac {\log (1 + x)}{1 + x^{2}}dx = \arctan x\cdot\log(1 + x)\Bigg|{0}^{1} - \int{0}^{1}\frac {\arctan x}{1 + x}dx=\frac{\pi\log2}{8}$$ which is answered here.

    https://math.stackexchange.com/questions/334075/evaluation-of-integral-int-01-frac-arctan-x-1x-dx

     – james givens Aug 18 '17 at 06:59
  • @K.Kwai:I have updated my answer to reflect this. – Paramanand Singh Aug 18 '17 at 07:50
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This is not an answer but it is too long for a comment.

I am not sure that we could get a "nice" expression for the infinite limit.

Looking at the inner sum $$\sum_{k=1}^{2n+1}\frac{(-1)^k} k =-\Phi (-1,1,2 n+2)-\log (2)$$ where appears the Lerch transcendent function. So, looking at the sum $$S=\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\left\{\sum_{k=1}^{2n+1}\frac{(-1)^k} k \right\}=-\sum _{n=0}^{\infty } \frac{(-1)^n \Phi (-1,1,2 n+2)}{2 n+1}-\log(2)\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}$$ The second summation is just $\frac \pi 4$ but for the first one, I (a CAS neither) found any result.

The only thing I was able to do is to compute the partial sums $$T_p=\sum _{n=0}^{p } \frac{(-1)^n \Phi (-1,1,2 n+2)}{2 n+1}$$ and to notice (as one would expect) oscillations and a "rather" slow convergence. Some values are reported below $$\left( \begin{array}{cc} p & T_p \\ 0 & 0.3068528194 \\ 1 & 0.2601241018 \\ 2 & 0.2781613324 \\ 3 & 0.2686789568 \\ 4 & 0.2745109280 \\ 5 & 0.2705657615 \\ 6 & 0.2734108826 \\ 7 & 0.2712625713 \\ 8 & 0.2729418771 \\ 9 & 0.2715932338 \\ 10 & 0.2727000562 \\ 11 & 0.2717754047 \\ 12 & 0.2725594174 \\ 13 & 0.2718862390 \\ 14 & 0.2724705249 \\ 15 & 0.2719586210 \\ 16 & 0.2724108044 \\ 17 & 0.2720084696 \\ 18 & 0.2723687660 \\ 19 & 0.2720442480 \\ 20 & 0.2723380637 \end{array} \right)$$ All of that seems to show that the infinite summation converges to something like $-0.816595$ which is not identified by inverse symbolic calculators.

Claude Leibovici
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