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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.

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

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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
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