TITLE: Can the Cauchy product of divergent series with itself be convergent?
QUESTION [14 upvotes]: For series $\sum a_n$ and $\sum b_n$ their Cauchy product is the series $\sum c_n$ where 
$$
c_n = a_0b_n+a_1b_{n-1}+\ldots+a_nb_0.
$$
Does there exist a sequence $\sum a_n$ such that:

$\sum a_n$ is divergent?
The Cauchy product of $ \sum a_n$ and $\sum a_n$ is convergent?

REPLY [6 votes]: Let $f(x)=\sum_{n=0}^\infty a_nx^n$ be the (binomial) Taylor series for the function $\sqrt{1+x}$. Then:
1) This power series has radius of convergence $1$, so the series $f(2)=\sum_{n=0}^\infty a_n2^n$ diverges.
2) On the interval $(-1,1)$, we have $f(x)f(x)=1+x$ pointwise, and so the Cauchy product of $f(x)$ with itself must be the formal power series $1+x+0x^2+0x^3+\dots$. In particular, this means that the Cauchy product of $f(2)$ with itself is the series $1+2+0+0+\dots$.
(This is essentially the same example as in all the other answers, but I feel like this is an easier way to come to it. We could replace $1+x$ with any other holomorphic function whose square root is not holomorphic on its entire domain.)