find the value of $ S = \sum _{k=1}^{\infty }\left(\frac{\left(-1\right)^{k+1}}{(k)(2k+1)}\right)$

Question

Find the value of $\frac{1}{3}-\frac{1}{10}+\frac{1}{21}-\frac{1}{36}+...$

I have no idea how to solve this infinite sum, I appreciate any help, thanks in advance.

Answer 1

$$\frac S2=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{2k(2k+1)} =\sum_{k=1}^\infty(-1)^{k+1}\left(\frac1{2k}-\frac1{2k+1}\right) =\sum_{k=1}^\infty(-1)^{k+1}\int_0^1(x^{2k-1}-x^{2k})\,dx =\int_0^1 \frac{x-x^2}{1+x^2}\,dx$$ etc.

Answer 2

Observe you have \begin{align} \operatorname{Log}(1+i)=\sum^\infty_{n=1} (-1)^n\frac{i^n}{n}=\sum^\infty_{k=1}\frac{(-1)^k}{2k}+i\sum^\infty_{k=1}\frac{(-1)^{k+1}}{2k-1} \end{align} where $\operatorname{Log}z$ is the principal branch. Hence it follows \begin{align} \operatorname{Log}(1+i)=\log 2+ i\arg (1+i) = \log 2 + i\frac{\pi}{4} \end{align} which means \begin{align} \sum^\infty_{k=1}\frac{(-1)^{k+1}}{2k+1}=1-\frac{\pi}{4}. \end{align}

For the other problem, we see that \begin{align} \sum^\infty_{k=1}(-1)^{k+1} \frac{1}{2k(2k+1)} = \sum^\infty_{k=1}(-1)^{k+1}\left( \frac{1}{2k}-\frac{1}{2k+1}\right)=-\frac{1}{2}\sum^\infty_{k=1}\frac{(-1)^k}{k}-\frac{\pi}{4} = \frac{1}{2}\log 2+\frac{\pi}{4}-1. \end{align}