(related to Proposition: Cauchy Product of Convergent Series Is Not Necessarily Convergent)
It is sufficient to find only one counter-example of two convergent series \(\sum_{n=0}^\infty a_n\) and \(\sum_{n=0}^\infty b_n\), the Cauchy product of which does not converge.
We construct such a counter-example. For a fixed \(n\) set
\[a_n:=b_n:=\frac{(-1)^n}{\sqrt{n+1}}\quad\text{and}\quad c_n:=\sum_{k=0}^n a_{n-k}b_k.\]
We will first show that the series \(\sum_{n=0}^\infty a_n\) and \(\sum_{n=0}^\infty b_n\) are convergent. This follows from the criterion for alternating infinite series, because the sequence \[\left(\frac{1}{\sqrt{n+1}}\right)_{n\in\mathbb N}\] is a monotonically decreasing real sequence, which converges to \(0\).
The Cauchy product has the terms
\[c_n:=\sum_{k=0}^n a_{n-k}b_k=\sum_{k=0}^n \frac{(-1)^{n-k}}{\sqrt{n-k+1}}\cdot\frac{(-1)^n}{\sqrt{n+1}}=(-1)^n\sum_{k=0}^n\frac 1{\sqrt{(n-k+1)(k+1)}}.\]
Now, for all \(k=0,\ldots,n\) we can estimate
\[(n-k+1)(k+1)=(n-k)k+(n-k)+(k+1)=(n-k)k+(n+1)\le n^2+(n+1) < n^2+2n+1=(n+1)^2,\]
and
\[\sqrt{(n-k+1)(k+1)} < (n+1).\]
and thus, for all \(k=0,\ldots,n\), we get the estimation
\[\frac {1}{\sqrt{(n-k+1)(k+1)}}>\frac {1}{(n+1)}.\]
With this estimation, we can therefore finally estimate
\[|c_n|=\left|\sum_{k=0}^n\frac {1}{\sqrt{(n-k+1)(k+1)}}\right| > \left|\sum_{k=0}^n\frac {1}{(n+1)}\right|=(n+1)\cdot \frac {1}{(n+1)}=1.\]
This demonstrates that the series \(\sum_{n=0}^\infty c_n\) cannot converge, since the absolute value of every term is greater then \(1\). (If the series converged, then it would violate the necessary condition for convergent series requiring all terms to converge to \(0\).)