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Find the sum of series $$\sum_{n=0}^{+\infty}\frac{(-1)^n}{2(n+1)(2n+1)}$$ I have tried to use the telescoping method, but it seems that it can't reduce the problem.

$$\sum_{n=0}^{+\infty}\frac{(-1)^n}{2(n+1)(2n+1)}=\lim_{N\rightarrow+\infty}\sum_{n=0}^{N}\frac{(-1)^n}{(2n+2)(2n+1)}$$ $$=\lim_{N\rightarrow+\infty}\sum_{n=0}^{N}(-1)^n\left(\frac{1}{2n+1}-\frac{1}{2n+2}\right)$$ $$=\lim_{N\rightarrow+\infty}\left[\left(1-\frac{1}{2}\right)-\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+...+(-1)^N\left(\frac{1}{2N+1}-\frac{1}{2N+2}\right)\right]$$

Is it possible to find the sum of the last expanded series?

user300045
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    Yes, it is possible. This question was asked before some times but I dont found now a copy. – Masacroso Mar 11 '17 at 16:34
  • maybe the final result rings a bell? http://www.wolframalpha.com/input/?i=Sum%5B(-1)%5En%2F((n%2B1)(2n%2B1)),%7Bn,0,inf%7D%5D – tired Mar 11 '17 at 16:35
  • $$\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}=-\sum_{n=0}^{\infty}\frac{(-1)^n}{n}=-\log(2) $$

    and

    $$\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}=\frac{\pi}{4} $$

    by the famous Leibnitz formula

     – tired Mar 11 '17 at 16:38
  • btw. both sums are (conditionally) convergent so the explicit limiting procedure isn't really necessary – tired Mar 11 '17 at 16:41
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    The sum of this series is $$(\Re+\Im)\left(\sum_{n=1}^\infty\frac{i^n}n\right)=-(\Re+\Im)(\log(1-i))=-(\Re+\Im)(\tfrac12\log2-i\tfrac\pi4)=-\tfrac12\log2+\tfrac\pi4$$ – Did Mar 11 '17 at 16:43
  • @tired can you please explain or cite a source of the proof of the second summation equal to $\pi/4$ – Navin Mar 11 '17 at 18:59
  • you can get this result straight from the taylor expansion of the $\arctan$ function – tired Mar 11 '17 at 19:11
  • or: use $\sum_{n\geq0}(-1)^nx^{2n}=1/(1+x^2)$. Now integrate both sides from $0$ to $1$ – tired Mar 11 '17 at 19:13

1 Answers1

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

A good approach to evaluate this sum is to try to make geometric sums appear, because we know how to handle them. To do this we utilize $\int_{0}^{1} x^n \,dx=\frac{1}{n+1}$. We get,

$$=\sum_{n \geq 0} (-1)^n \int_{0}^{1} (x^{2n}-x^{2n+1}) \,dx$$

Interchange integral and sum.

$$=\int_{0}^{1} \sum_{n \geq 0} (-1)^{n} \left(x^{2n}-x^{2n+1}\right) \,dx$$

Factor out $x^{2n}$.

$$=\int_{0}^{1} \sum_{n \geq 0} (-1)^{n} \left(x^{2n}(1-x)\right) \,dx $$

Factor out $1-x$ from the sum because it does not depend on $n$

$$=\int_{0}^{1} (1-x) \sum_{n \geq 0} (-1)^n x^{2n} \,dx$$

$$=\int_{0}^{1} (1-x) \sum_{n \geq 0} (-1)^n (x^2)^n \,dx$$

$$=\int_{0}^{1} (1-x) \sum_{n \geq 0} (-x^2)^n \,dx$$

Recognize geometric sum.

$$=\int_{0}^{1} \frac{1-x}{1+x^2} dx$$

Split integral into two standard very integrals.

$$=\int_{0}^{1} \frac{1}{1+x^2} \,dx -\frac{1}{2} \int_{0}^{1} \frac{2x}{1+x^2} \,dx$$

$$=\frac{\pi}{4}-\frac{1}{2}\ln 2$$

Ahmed S. Attaalla
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