Convergent series whose Cauchy product diverges 
Give an example of convergent series $\sum_{n=0}^{\infty}a_n$ and $\sum_{n=0}^{\infty}b_n$ whose Cauchy product diverges.

So at least one series must not be absolutely convergent, since if both are absolutely convergent then the Cauchy product absolutely converges. I'm thinking about the sequence $\{a_n\} = 1, -1/2, 1/3, -1/4, \ldots$, but it seems hard to find the Cauchy product.

We will use as base series the slowly convergent series
$$\frac{1}{1^{1/4}} -\frac{1}{2^{1/4}}+\frac{1}{3^{1/4}}-\frac{1}{4^{1/4}}+\frac{1}{5^{1/4}}-\frac{1}{6^{1/4}}+\frac{1}{7^{1/4}}-\frac{1}{8^{1/4}}+\frac{1}{9^{1/4}}-\frac{1}{10^{1/4}}+\cdots.$$
So we are using the series $\sum_{i=1}^\infty (-1)^{i+1} \frac{1}{i^{1/4}}$. 
(It is convenient to start the indexing at $i=1$. But if you really want to start at $0$, prepend an additional $9$-th tern of $0$.) Consider the Cauchy product of the above series with itself. 
In particular, consider the term $c_{n+1}$ of the Cauchy product, where for convenience $n$ is odd. For a concrete example, let $n=9$.  Then $c_{n+1}$ is a sum of $n$ terms. Each term is $\ge \frac{1}{\left(\frac{n+1}{2}\right)^{1/2}}$. This is because for any positive integer $a$, the product $x(a-x)$ attains its maximum at $x=\frac{a}{2}$.  
It follows that the sequence $(c_{n})$ does not have limit $0$, so $\sum c_n$ does not converge.  
