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:
Let $c_0=1$, $c_1=-2$, $c_n=0$ for $n>1$. Let $a_0=1$, and for $n\ge1$ recursively $$a_n=\frac12\left(c_n-\sum_{k=1}^ {n-1}a_ka_{n-k} \right).$$ Then clearly $\sum c_n=\sum a_n\cdot \sum a_n$ in the sense of Cauchy product, and $\sum c_n$ is of course very convergent. Assume $\sum a_n$ converges. Then $\sum a_n x^n$ converges absolutely on $(-1,1)$. By absolute convergence, $\sum c_n x^n=(\sum a_n x^n)^2$ for $|x|<1$, which is absurd as $\sum c_n x^n<0$ for $x>\frac12$.
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.)
Without lost of generality, assume $a_0 \neq 0$. One proves that the Cauchy relations $c_n = \sum_{j=0}^n a_j a_{n-j}$ are equivalent to $$ a_n = \frac{1}{2a_0}\left(c_n - \sum_{j=1}^{n-1} a_j a_{n-j} \right) \, .$$ Multiplying $\sum_n a_n$ by an overall constant if necessary, we can assume that $2a_0 = 1$. It follows that $c_0 = 1/4$ and that $c_1 = a_1$.
Let's consider the case when $c_n = 0$ for $n \ge 2$, so that $\sum_n c_n = 1/4 + c_1$ is convergent. It is easily proved by induction on $n \ge 1$ that there exist positive integers $A_n$ such that $a_n = (-1)^{n-1}A_n c_1^n$. Therefore, whenever $|c_1| \ge 1$, the limit $\lim_{n\to \infty} a_n$ is not $0$ (if it exists at all), and so the series $\sum_n a_n$ diverges.
Remark: In fact, based on John's answer, $A_{n+1} = C_n = \frac{1}{n+1} {2n \choose n}$ is the $n$-th Catalan number; Stirling's approximation allows to estimate $A_{n+1} \sim \frac{4^n}{\sqrt{\pi} n^{1/2}(n+1)}$, so that the above construction works if (and only if) $|c_1| > 1/4$.
I managed to find a construction.
Put $a_0 = \frac{1}{2}$, $a_n = (-1)^{n-1}C_{n-1}x^n$ for $n=1,2,3,...$ where $C_n$ is the n-th Catalan number.
So that $\sum a_n = \frac{1}{2} + x -x^2+2x^3-5x^4+14x^5-42x^6+...$
For $n \geq 2 $ we have:
$c_n = [(-1)^{n-1}C_{n-1} - (-1)^{n-2}(C_0C_{n-2} + C_1C_{n-3}+...+C_{n-2}C_0)]\cdot x^n$
Using the known formula: $C_n = C_0C_{n-1}+C_1C_{n-2}+...+C_{n-1}C_0$
We have $c_n$ =0
So $\sum c_n = \frac{1}{4} + x + 0 +0 + ...$ which obviously converges. Now putting $x=1$ we found the required series. ($\sum a_n$ diverges because $C_n$ doesn't tend to 0 as $n \to \infty$)
