<s>[INST] find the value of $ S = \sum _{k=1}^{\infty }\left(\frac{\left(-1\right)^{k+1}}{(k)(2k+1)}\right)$ 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. 
 [/INST] 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}
</s>