If $a_{n} = ((-1)^{n})/(\sqrt{1+n})$ then is $\sum_{k=0}^{n} (a_{n-k})(a_{k})$ convergent? If $a_{n} = ((-1)^{n})/(\sqrt{1+n})$ then is $C_{n} = \sum_{k=0}^{n} (a_{n-k})(a_{k})$ convergent?
$a_{n}$ is conditionally convergent, but how do I check the convergence of $C_{n}$?
I could transform $C_{n}$ in following way 
$C_{n} = \sum_{k=0}^{n} \frac{(-1)^{n}}{(1+k)^{1/2}(1+n-k)^{1/2}}$ 
How to proceed?

$C_{n} 
= \sum_{k=0}^{n} \frac{(-1)^{n}}{(1+k)^{1/2}(1+n-k)^{1/2}}
$
$\begin{array}\\
(1+k)(1+n-k)
&=1+n+kn-k^2\\
&=1+n-(k^2-kn)\\
&=1+n-(k^2-kn+n^2/4)+n^2/4\\
&=1+n-(k-n/2)^2+n^2/4\\
&= (n+2)^2/4-(k-n/2)^2\\
\text{so}\\
\dfrac1{\sqrt{(1+k)(1+n-k)}}
&= \dfrac1{\sqrt{(n+2)^2/4-(k-n/2)^2}}\\
\end{array}
$
so
$\begin{array}\\
|C_{n}| 
&= \sum_{k=0}^{n} \dfrac{1}{\sqrt{(n+2)^2/4-(k-n/2)^2}}\\
&= \sum_{k=0}^{n} \dfrac{1}{n\sqrt{(1+2/n)^2/4-(k/n-1/2)^2}}\\
&\approx \int_0^1 \dfrac{dx}{\sqrt{1/4-(x-1/2)^2}}
\quad\text{Riemann sum}\\
&= \pi
\quad\text{since }\int \dfrac{dx}{\sqrt{1/4-x^2}}=\arcsin(2x)\\
\end{array}
$
Wolfy says that
$(n, |C_n|)
=(10, 2.293...), (100, 2.852...), (1000, 3.049...), (10000, 3.1123...)
$
so the convergence is there
but slow.
