If $a_n=b_n = \frac{(-1)^n}{n+1}$, how to deduce that the Cauchy product series converges? If $c_n=\sum_{k=0}^{n}a_kb_{n-k}$, we have to prove that,
$$c_n=(-1)^n\frac{2}{j+2}\left(1+\frac{1}{2}+\cdots+\frac{1}{k+1}\right)$$
But what I get is:
$$c_n=(-1)^n\sum_{k=0}^{n}\frac{1}{(k+1)((n-k)+1)}=(-1)^n\left(\frac{1}{1(n-(-1))}+\frac{1}{2(n-0)}+\frac{1}{3(n-1)}+\cdots+\frac{1}{(n+1)(n-(n-1))}\right)$$
How to proceed further? Thanks in advance.

We have
$$\tag{*}\sum_{n=0}^\infty c_n = \sum_{n=0}^\infty (-1)^n\underbrace{\sum_{k=0}^n \frac{1}{(k+1)(n-k+1)}}_{d_n},$$
where $d_n$ can be expressed in terms of the harmonic sum $H_{n+1} = \sum_{k=1}^{n+1} \frac{1}{k}$, viz.
$$d_n =\sum_{k=0}^n \frac{1}{(k+1)(n-k+1)}= \sum_{k=0}^n\frac{1}{n+2} \left(\frac{1}{k+1} + \frac{1}{n+1-k} \right) \\= \frac{1}{n+2} \left(\sum_{k=0}^n \frac{1}{k+1} + \sum_{k=0}^n \frac{1}{n+1-k} \right) = \frac{2}{n+2}\sum_{k=0}^n\frac{1}{k+1} = \frac{2H_{n+1}}{n+2}$$
It is straightforward to show that $d_n$ is decreasing and convergent to $0$ since $H_{n+1} \sim \log (n+1)$ as $n \to \infty$.
Thus, (*) is convergent by the alternating series test.
