# Series counterexample

Give an example of series

$$\sum_{n=1}^{\infty} a_n \quad \text{ and } \quad \sum_{n=1}^{\infty} b_n$$

such that both converge, and the series

$$\sum_{n=1}^{\infty} c_n$$

defined by

$$c_n = a_{n-1}b_1 + a_{n-2}b_2 + \cdots + a_1b_{n-1}$$

diverges.

-

The canonical example is $$a_n=b_n=\frac{(-1)^n}{\sqrt n}$$
ADD Given two sequences $a_n,b_n$, the sequence $c_k=\sum_{i=1}^k a_ib_{k-i}$ is usually called the Cauchy product or convolution of $a_n$ with $b_n$. It is a good exercise (and not an easy one) to prove that if $a_n$ is absolutely summable - that is $$\sum |a_n|$$ exists - and $b_n$ is summable, then the Cauchy product is summable and it converges to the product of the sums. This is known as Merten's theorem.
ADD There is a theorem (found in Spivak's calculus) that says that if both $a_n$ and $b_n$ are absolutely summable, then any sum of the form $$\sum_{i,j} c_{i,j}$$ where each product $a_\ell b_k$ appears exactly once will converge to $$\sum a_n\cdot \sum b_n$$
+1. Note that it is sufficient to have $\sum a_n$ absolutely convergent and $\sum b_n$ convergent (not necessarily absolutely so) for the Cauchy product to be convergent to the product of the sums. – Ayman Hourieh Mar 22 '13 at 23:32
Is $a_n=b_n=\dfrac {(-1)^n}{n^{\dfrac {1}{k}}}$ a counterexample for every $k\geq2$? – user67878 Mar 22 '13 at 23:43
@Thus ... and then try for $1 \lt k \lt 2$ – Henry Mar 23 '13 at 0:09