Originally Posted by

**Dinkydoe** A part of my course complex-analysis we deal with infinite products. And as a homework assignment I got the following exercise.

If $\displaystyle b_n>1$ for all $\displaystyle n$, then prove that $\displaystyle \prod_n b_n$ converges iff $\displaystyle \sum_{n}\log(b_n) < \infty $

I'm used to not so easy homework-assignments, so that's why I feel I'm missing something here.

My answer would be:

$\displaystyle \Rightarrow$

if $\displaystyle \prod_n b_n=M$ we have $\displaystyle M\geq 1$ since $\displaystyle b_n>1$ for all $\displaystyle n$. Hence $\displaystyle \log(M)<\infty$ and thus we obtain $\displaystyle \log(M)=\log(\prod_n b_n)= \sum_n \log(b_n) <\infty$

$\displaystyle \Leftarrow $

If $\displaystyle \sum_n \log(b_n) = \log(\prod_n b_n) = M$ we have $\displaystyle e^M= \prod_n b_n<\infty $

So am I missing something?