Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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


share|improve this question

1 Answer 1

up vote 10 down vote accepted

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 $$\sum a_n\cdot \sum b_n$$

share|improve this answer
Great bonus question. –  nkhuyu Mar 22 '13 at 23:29
+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 Try and prove it! =) (or disprove it) –  Pedro Tamaroff Mar 22 '13 at 23:59
@Thus ... and then try for $1 \lt k \lt 2$ –  Henry Mar 23 '13 at 0:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.