What is an elegant case of two convergent series $\sum a_n$ and $\sum b_n$ which Cauchy product
$$\sum_{n=0}^{\infty} c_n = \sum_{n=0}^\infty \sum_{k=0}^n a_kb_{n-k}$$
diverges?
The most common, I believe, is $a_n=b_n=\frac{(-1)^n}{\sqrt{n+1}}$, in which case
$$c_n=\sum_{k=0}^n(-1)^n\frac{1}{\sqrt{k+1}\sqrt{n+1-k}}$$
and from the inequalities $k+1\leq n+1$ and $n+1-k\leq n+1$ it follows that the denominator is smaller than $n+1$, and
$$|c_n|\geq \sum _{k=0}^n \frac{1}{n+1}=1$$
so the condition $c_n\rightarrow 0$, necessary for converge, does not hold.
Here is a family of counter examples to the Cauchy product convergence. Consider the conditional convergent series $\sum^\infty_{n=0}\tfrac{(-1)^{n+1}}{(n+1)^\alpha}$ and $\sum^\infty_{n=0}\frac{(-1)^{n+1}}{(n+1)^\beta}$ where $0<\alpha,\beta<1$. $$c_n:=\sum^n_{k=0}\frac{(-1)^{k+1}}{(k+1)^{\alpha}}\frac{(-1)^{n-k+1}}{(n-k+1)^\beta}=(-1)^n\sum^n_{k=0}\frac{1}{(k+1)^{\alpha}}\frac{1}{(n-k+1)^\beta}$$ As $1\leq k+1,n-k+1\leq n+1$ whenever $0\leq k\leq n$, $$1\leq (k+1)^\alpha(n-k+1)^\beta\leq (n+1)^{\alpha+\beta}$$ If $\alpha+\beta\leq1$ $$|c_n|\geq\sum^n_{k=0}\frac{1}{(n+1)^{\alpha+\beta}}=(n+1)^{1-\alpha-\beta}\geq1$$ Hence $\sum_nc_n$ is a divergent series.
More examples like the one above (alternating series with monotone decreasing driver) can be obtained. Consider $\sum^\infty_{n=0}(-1)^na_n$ and $\sum^\infty_{n=0}(-1)^nb_n$ where $a_n$ and $b_n$ are decreasing sequences converging to $0$. $$c_n:=\sum^n_{k=0}(-1)^ka_k(-1)^{n-k}b_{n-k}=(-1)^n\sum^n_{k=0}a_kb_{n-k}$$ $a_n\leq a_k\leq a_0$ and $b_n\leq b_k\leq b_0$ for all $0\leq k\leq n$. Thus $$|c_n|\geq \sum^n_{k=0}a_nb_n=(n+1)a_nb_n$$ Choosing sequences $a_n$ and $b_n$ so that $na_nb_n$ does not converge to $0$ will give a plethora of counter examples.
Here is an example of convergence of Cauchy product of two conditionally convergent series. Consider the series $\sum^\infty_{n=0}\frac{(-1)^{n+1}}{n+1}$. Then $$\begin{align} \sum^n_{k=0}\frac{(-1)^{k+1}}{k+1}\frac{(-1)^{n-k+1}}{n-k+1}&=(-1)^n\sum^n_{k=0}\frac{1}{k+1}\frac{1}{n-k+1} \end{align}$$ Since $$\frac{1}{k+1}\frac{1}{n-k+1}=\frac{1}{n+2}\Big(\frac{1}{k+1}+\frac{1}{n-k+1}\Big)$$ $$\begin{align} \sum^n_{k=0}\frac{(-1)^{k+1}}{k+1}\frac{(-1)^{n-k+1}}{n-k+1}&=\frac{2(-1)^n}{n+2}\sum^n_{k=0}\frac{1}{k+1} \end{align}$$ Hence, the Cauchy product of $\sum^{\infty}_{n=0}\frac{(-1)^n}{n+1}$ with itself is $$ \sum^\infty_{n=1}\frac{(-1)^n}{n+1}H_n $$ where $H_n=\sum^n_{k=1}\frac{1}{k}$ is the $n$-th harmonic sum. It is left to the OP to show that the latter series converges ($H_n\sim\log n$)
There are situations where the Cauchy product of two divergent series converge