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?

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?

Last edited: