To show that the Cauchy product diverges $$
\sum\frac{(-1)^n}{\sqrt{n+1}}
$$
$a)$ Show that this series converges- it's easy to show that $\lim_{n \to \infty} \frac{(-1)^n}{\sqrt{n+1}} = 0 $
$b)$Show that the Cauchy Product with itself diverges. 
Edit: According to my book the Cauchy product of two sequences $\sum_{n \in \mathbb{N}} a_n$ and  $\sum_{n \in \mathbb{N}} b_n$ is defined by $\sum_{n \in \mathbb{N}} c_n$ where-
$$
c_n = \sum_{k=0}^{n} a_{n-k} b_{k}
$$
 A: If $a_n = b_n = (-1)^n/\sqrt{n+1}$, then the Cauchy product is
$$\sum_{n=0}^{\infty}c_n = \sum_{n=0}^{\infty}\sum_{k=0}^{n}a_kb_{n-k}.$$
Note that 
$$c_n = \sum_{k=0}^{n}a_kb_{n-k}= \sum_{k=0}^{n}\frac{(-1)^k}{\sqrt{k+1}}\frac{(-1)^{n-k}}{\sqrt{n-k+1}}\\= (-1)^n\sum_{k=0}^{n}\frac{1}{\sqrt{k+1}}\frac{1}{\sqrt{n-k+1}}.$$
Then 
$$|c_n| = \sum_{k=0}^{n}\frac{1}{\sqrt{k+1}}\frac{1}{\sqrt{n-k+1}}\geq \sum_{k=0}^{n}\frac{1}{\sqrt{n+1}}\frac{1}{\sqrt{n+1}}= 1.$$
Since $c_n$ does not converge to $0$, $\sum c_n$ diverges.
Note that 
$$\frac{1}{\sqrt{k+1}}\frac{1}{\sqrt{n-k+1}} \geq \frac{1}{\sqrt{n+1}}\frac{1}{\sqrt{n+1}}= \frac1{n+1}$$ can be shown as follows:
$$\sqrt{k+1}\sqrt{n-k+1} = \sqrt{nk-k^2+n+1} \leq \sqrt{n^2 + 2n +1}= n+1.$$
A: The given series converges by Leibniz Test. About its Cauchy product with itself: the general term (even index) of that product is
$$c_{2n}=\sum_{k=0}^{2n}\frac{(-1)^k}{\sqrt{k+1}}\frac{(-1)^{2n-k}}{\sqrt{2n-k+1}}=$$
$$=\frac1{\sqrt{2n+1}}+\frac1{\sqrt2\sqrt{2n}}+\frac1{\sqrt3\sqrt{2n-1}}+\ldots\frac1{\sqrt{2n+1}}=$$
$$2\left(\frac1{\sqrt{2n+1}}+\ldots+\frac1{\sqrt n\sqrt{n+1}}\right)\ge2\frac n{\sqrt n\sqrt{n+1}}\rlap{\;\;\;\;\;/}\xrightarrow[n\to\infty]{}0$$
