Need help to check alternating series criterion I know this is simple calculation based question but I got stuck:
I want to check convergence of the following series:
$$1-\frac{1}{2}(1+\frac{1}{3})+\frac{1}{3}(1+\frac{1}{3}+\frac{1}{5})-\cdots$$
Clearly this is alternating series. So I need satisfy alternating series criterion. That means I need to show that its terms are decreasing monotonically.
Let $a_n=\frac{1}{n}(1+ \frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n-1})$ and  $a_{n+1}=\frac{1}{n+1}(1+ \frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n+1})$.
I just have to show  $a_n-a_{n+1}>0$  but right here I got stuck because of the presence of the first factor. I think i am missing a simple trick here.
Anyway, I got $$a_n-a_{n+1}=(\frac{1}{n}-\frac{1}{n+1})+\frac{1}{3}(\frac{1}{n}-\frac{1}{n+1})+\cdots+\frac{1}{2n-1}(\frac{1}{n}-\frac{1}{n+1}) {-\color{red}{\frac{1}{(2n+1)(n+1)}}}.$$
The last red color term making problem.
May be ratio test help as well ?
Any help please
 A: $\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n(n+1)}\gt\frac{1}{(2n+1)(n+1)}$.  Therefore that extra term is taken care of.
A: Doesn't answer your question, since I think you've already got the answer to it. It ads something to it though.
Let's evaluate the sum of the series $ \sum\limits_{n\geq 0}{\frac{\left(-1\right)^{n}}{n+1}\sum\limits_{k=0}^{n}{\frac{1}{2k+1}}} $.
Denoting $\left(\forall n\in\mathbb{N}\right),\ a_{n}=\frac{\left(-1\right)^{n}}{n+1}\sum\limits_{k=0}^{n}{\frac{1}{2k+1}} $, and $ b_{n}=\frac{\left(-1\right)^{n}}{2n+1} $.
Let $ n\in\mathbb{N} $, we have the following : $$ \sum_{k=0}^{n}{\frac{1}{2k+1}}=\frac{1}{2}\left(\sum_{k=0}^{n}{\frac{1}{2k+1}}+\sum_{k=0}^{n}{\frac{1}{2\left(n-k\right)+1}}\right) $$
We added the sum to itsefl, but by shifting the terms, then we divided by $ 2 $.
Thus : $$ \sum_{k=0}^{n}{\frac{1}{2k+1}}=\left(n+1\right)\sum_{k=0}^{n}{\frac{1}{\left(2k+1\right)\left(2\left(n-k\right)+1\right)}} $$
Hence : $$ a_{n}=\sum_{k=0}^{n}{\left(\frac{\left(-1\right)^{k}}{2k+1}\times\frac{\left(-1\right)^{n-k}}{2\left(n-k\right)+1}\right)} $$
Now, using a Cauchy product, we have : \begin{aligned}\sum_{n=0}^{+\infty}{\frac{\left(-1\right)^{n}}{n+1}\sum_{k=0}^{n}{\frac{1}{2k+1}}}&=\sum_{n=0}^{+\infty}{a_{n}}\\&=\sum_{n=0}^{+\infty}{\sum_{k=0}^{n}{b_{k}b_{n-k}}}\\ &=\left(\sum_{n=0}^{+\infty}{b_{n}}\right)^{2}\\ &=\left(\sum_{k=0}^{n}{\frac{\left(-1\right)^{n}}{2n+1}}\right)^{2}\\ \sum_{n=0}^{+\infty}{\frac{\left(-1\right)^{n}}{n+1}\sum_{k=0}^{n}{\frac{1}{2k+1}}}&=\frac{\pi^{2}}{4}\end{aligned}
