# If $a_n=\frac{(-1)^n}{\sqrt{1+n}}$ and $c_n=\sum_{k=0}^n a_{n-k}a_k$, does $\sum_{n=0}^\infty c_n$ converge?

Suppose for $$n \in \mathbb{N} \cup \{0\}$$, $$a_n=\frac{(-1)^n}{\sqrt{1+n}}$$. Define $$c_n=\sum_{k=0}^na_{n-k}a_k$$. Does $$\sum_{n=1}^\infty c_n$$ converge?

What I attempted:- \begin{aligned} c_n&=\sum_{k=0}^na_{n-k}a_k \\ &=a_na_0+a_{n-1}a_1+\dots+a_0a_n\\ &=\frac{(-1)^n}{\sqrt{1+n}}\times 1+\frac{(-1)^{n-1}}{\sqrt{1+n-1}}\times \frac{(-1)}{\sqrt{1+1}}+\frac{(-1)^{n-2}}{\sqrt{1+n-2}}\times \frac{(-1)^2}{\sqrt{1+2}}\dots+1 \times \frac{(-1)^n}{\sqrt{1+n}} \\ &=(-1)^n\left(\frac{1}{\sqrt{1+n}}+\frac{1}{\sqrt{2n}}+\frac{1}{\sqrt{3(n-1)}}+\dots+\frac{1}{\sqrt{1+n}}\right)\\ &=(-1)^nd_n \end{aligned} $$(d_n)_{n=1}^\infty$$ is a decreasing sequence and $$\lim_{n \to \infty}d_n=0$$. Hence $$\sum_{n=0}^{\infty}(-1)^nd_n=\sum_{n=0}^{\infty}c_n$$ converges. As a result $$\sum_{n=1}^{\infty}c_n$$ also converges.

Am I correct?

Hint. Note that $$d_n=\sum_{k=0}^n\frac{1}{\sqrt{(1+k)(n+1-k)}}\geq \sum_{k=0}^n\frac{2}{(1+k)+(n+1-k)}=\frac{2(n+1)}{n+2},$$ and therefore $$c_n=(-1)^n d_n$$ does not tend to zero as $$n$$ goes to infinity.
Your reasoning is correct, but the claim that $$d_n$$ is decreasing and $$\lim_{n\to\infty} d_n = 0$$ needs to be proven, not just stated. (In fact, it is not actually true, although it may look true.)
If $$a_n=\frac{(-1)^n}{\sqrt{n+1}}$$ then $$f(x)=\sum_{n\geq 0}a_n x^n$$ has a singularity of the $$\frac{1}{\sqrt{x+1}}$$ kind at $$x=-1$$ and $$g(x)=\sum_{n\geq 0}c_n x^n = f(x)^2$$ has a simple pole at $$x=-1$$. In particular $$|c_n|$$ does not converge to zero as $$n\to +\infty$$ and $$\sum_{n\geq 1}c_n$$ is not convergent. On the other hand, $$\sum_{n\geq 1}c_n$$ is convergent "à-la-Cesàro", like $$\sum_{n\geq 1}(-1)^n$$, i.e. the sequence of the averaged partial sums is convergent.
The situation is much clearer if $$a_n$$ is replaced by $$\frac{\sqrt{\pi}(-1)^n}{4^n}\binom{2n}{n}$$: in such a case $$f(x)=\sqrt{\frac{\pi}{1+x}}$$ and in the Cesàro sense we have $$\sum_{n\geq 1}c_n=\frac{\pi}{2}$$.