# Convergent sum / divergent product

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• Nov 14th 2011, 07:21 PM
masnarski
Convergent sum / divergent product
Hi guys,

This is apparently a fairly well-known example, but I can't seem to figure it out. Anyone know this? Want

$\sum_{n=1}^\infty a_n < \infty$

but

$\prod_{n=1}^\infty (1+a_n) \rightarrow \infty$

Now, we know that the sum will have to be conditionally convergent, as
$\sum_{n=1}^\infty |a_n| < \infty \Rightarrow \prod_{n=1}^\infty (1+a_n) < \infty$
and moreover, it is a theorem that if the sequence is square-summable, i.e.
$\sum_{n=1}^\infty |a_n|^2 < \infty$
then the product and the sum (without absolute values) converge and diverge together.

The sequences
$a_n = (-3)^n$
$a_n = \frac{(-1)^n}{n}$
$a_n = \frac{(-1)^n}{\sqrt{n}}$

don't seem to work, although I might be wrong about this last one. what about

$a_n = \frac{i\cdot(-1)^n}{\sqrt{n}}$
• Nov 14th 2011, 08:07 PM
masnarski
Re: Convergent sum / divergent product
Seems I've answered my own question as

$a_n = i^n/\sqrt{n}$ works!