Divergence of $\sum_{k=0}^n\frac1{\sqrt{(n-k+1)(k+1)}}$ This is exercise 2.3.42 in Sohrab's Basic Real Analysis:
Problem
Consider the alternating series $\sum_{n=0}^\infty(-1)^n/ \sqrt{n+1}$, which is convergent (why?). Show that the (Cauchy) product of this series with itself is $\sum_{n=0}^\infty z_n$, where
$
z_n=(-1)^n\sum_{k=0}^n\frac{1}{\sqrt{(n-k+1)(k+1)}},
$
and that this series is divergent.
Hint: $(n-k+1)(k+1)=(n/2+1)^2-(n/2-k)^2$.
Question
I want to prove the divergence of $\sum_{k=0}^n\frac{1}{\sqrt{(n-k+1)(k+1)}}$ (this is not necessary to solve the given problem as the answer of Parcly Taxel shows). I see that the series is something like $\sum 1/k$, but I was not able to show it formally. 
 A: Numerically it seems that the series does not diverge, it seems to converge to $\pi$ very slowly (by increasing $n$ of a factor $100$, an additional digit of $\pi$ is added). I will consider opening a question to prove the convergence.
EDIT A nicer form:
Let us consider $n=2k$ even, then 
$$
\sum_{k=0}^n\frac{1}{\sqrt{(n-k+1)(k+1)}}=\sum_{k=0}^n\frac{1}{\sqrt{(n/2+1)^2-(n/2-k)^2}}=
$$
$$
=2\sum_{k=0}^{n/2}\frac{1}{\sqrt{(n/2+1)^2-(n/2-k)^2}}=2\sum_{j=0}^{n/2}\frac{1}{\sqrt{(n/2+1)^2-j^2}}
$$
$$
=2\sum_{j=0}^{k}\frac{1}{\sqrt{(k+1)^2-j^2}}
$$
And, for $0<a\in\mathbb{R}$,
$$
\int_0^a\frac{dx}{\sqrt{a^2-x^2}}=\frac{\pi}{2}.
$$
A: For all $n\in\Bbb N$ and $0\le k\le n$ we have
$$(n-k+1)(k+1)=(n/2+1)^2-(n/2-k)^2\le(n/2+1)^2$$
Hence
$$\sqrt{(n-k+1)(k+1)}\le n/2+1$$
$$\frac1{\sqrt{(n-k+1)(k+1)}}\ge\frac1{n/2+1}$$
$$|z_n|=\sum_{k=0}^n\frac1{\sqrt{(n-k+1)(k+1)}}\ge\sum_{k=0}^n\frac1{n/2+1}=\frac{n+1}{n/2+1}$$
$\sum_{n=0}^\infty z_n$ is an alternating series. For it to be convergent, the limit of the absolute value of its terms must be zero, but
$$\lim_{n\to\infty}|z_n|=\lim_{n\to\infty}\sum_{k=0}^n\frac1{\sqrt{(n-k+1)(k+1)}}$$
$$\ge\lim_{n\to\infty}\frac{n+1}{n/2+1}=2\ne0$$
Hence the series diverges.
