prove the divergence of cauchy product of convergent series $a_{n}:=b_{n}:=\dfrac{(-1)^n}{\sqrt{n+1}}$ i am given these series which converge. $a_{n}:=b_{n}:=\dfrac{(-1)^n}{\sqrt{n+1}}$ i solved this with quotient test and came to $-1$, which is obviously wrong. because it must be $0<\theta<1$ so that the series converges. my steps: 
$\dfrac{(-1)^{n+1}}{\sqrt{n+2}}\cdot \dfrac{\sqrt{n+1}}{(-1)^{n}} = - \dfrac{\sqrt{n+1}}{\sqrt{n+2}} = - \dfrac{\sqrt{n+1}}{\sqrt{n+2}}\cdot \dfrac{\sqrt{n+2}}{\sqrt{n+2}} = - \dfrac{(n+1)\cdot (n+2)}{(n+2)\cdot (n+2)} = - \dfrac{n^2+3n+2}{n^2+4n+4} = -1 $
did i do something wrong somewhere?
and i tried to know whether the cauchy produkt diverges as task says: 
$\sum_{k=0}^{n}\dfrac{(-1)^{n-k}}{\sqrt{n-k+1}}\cdot \dfrac{(-1)^{k}}{\sqrt{k+1}} = \dfrac{(-1)^n}{nk+n-k^2+1} = ..help.. = diverging $
i am stuck here how to show that the produkt diverges, thanks for any help!

$\sum_{n=0}^\infty\dfrac{(-1)^n}{\sqrt{n+1}}$ is convergent by Leibniz's test, but it is not absolutely convergente (i.e. it is conditionally convergent.)
To show that the Cauchy product does not converge use the inequality
$$
x\,y\le\frac{x^2+y^2}{2}\quad x,y\in\mathbb{R}.
$$
Then
$$
\sqrt{n-k+1}\,\sqrt{k+1}\le\frac{n+2}{2}
$$
and
$$
\sum_{k=0}^n\frac{1}{\sqrt{n-k+1}\,\sqrt{k+1}}\ge\frac{2(n+1)}{n+2}.
$$
This shows that the the terms of the Cauchy product do not converge to $0$, and the series diverges.
