TITLE: Can we take a logarithm of an infinite product?
QUESTION [7 upvotes]: Suppose we have an infinite product $S = \prod_{n=1}^{\infty} a_n$ of positive real numbers. Then is it always the case that
$$
\log(S) = \sum_{n=1}^{\infty} \log a_n ?
$$
I am sure this is the case, but I wanted to make sure. 
Thank you!

REPLY [8 votes]: \begin{align}
\log\prod_{n=1}^\infty a_n&=\log\lim_{k\to\infty}\prod_{n=1}^ka_n\\
&=\lim_{k\to\infty}\log\prod_{n=1}^ka_n\\
&=\lim_{k\to\infty}\sum_{n=1}^k\log a_n\\
&=\sum_{n=1}^\infty\log a_n
\end{align}
I was able to switch the $\log$ and the $\lim$ because $\log$ is continuous.

REPLY [4 votes]: Yes (with a small caveat on whether you want to deal with $-\infty$ as a sum). If the product converges to some $S > 0$, then
$$\ln \prod_{n=1}^N a_n\xrightarrow[N\to\infty]{} \ln S$$
by continuity of the logarithm. But we do have
$$
\ln \prod_{n=1}^N a_n = \sum_{n=1}^N \ln a_n
$$
so we do have that the series $\sum_{n=1}^N \ln a_n$ is convergent, and its limit is indeed $\ln S$.
Now, if $S=0$, you do have $\sum_{n=1}^N \ln a_n \xrightarrow[N\to\infty]{} -\infty$, but it's up to you whether you want to call this "$\ln S$"...