Let $a_n =\frac {(-1)^n}{(1+n)^{0.5}}$ and let $c_n = ∑a_k a_{n-k}$
Then will $\sum^\infty _{n=0} c_n$ be converge?
If $\sum^\infty _{n=0} a_n$ converged absolutely I could say that the sequence $\sum^\infty _{n=0} c_n$ .
But in this case I have no idea how to solve it.
We have that
$$c_n = \sum_{k=0}^n a_k a_{n-k}=c_n = \sum_{k=0}^n \frac {(-1)^k(-1)^{n-k}}{(1+k)^{0.5}(1+n-k)^{0.5}}= \sum_{k=0}^n \frac {(-1)^n}{(1+n+nk-k^2)^{0.5}}\not \to 0$$
indeed for $n$ even
$$\sum_{j=-\frac n2}^{\frac n2} \frac {1}{(1+n+n(j+n/2)-(j+n/2)^2)^{0.5}}=\sum_{j=-\frac n2}^{\frac n2} \frac {1}{(1+n+nj+n^2/2-j^2-nj-n^2/4)^{0.5}}=\sum_{j=-\frac n2}^{\frac n2} \frac {1}{(1+n+n^2/4-j^2)^{0.5}}=2\sum_{j=0}^{\frac n2} \frac {1}{(1+n+n^2/4-j^2)^{0.5}}\neq 0$$
and therefore $\sum^\infty _{n=0} c_n$ does not converge.