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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{equation} \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} \end{equation} $(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?

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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.

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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.)

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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 $\sum_{n\geq 1}c_n=\frac{\pi}{2}$.

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