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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!

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    $\begingroup$ If the infinite product converges to a positive number, then continuity of the logarithm function permits the interchange of the limit and log. $\endgroup$ – Mark Viola Feb 2 '16 at 20:55
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\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.

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    $\begingroup$ It's may seem like splitting hairs, but writing $\ln \prod_{n=1}^\infty a_n$ when $\prod_{n=1}^\infty a_n=0$ itches a bit. :) $\endgroup$ – Clement C. Feb 2 '16 at 21:06
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    $\begingroup$ Yeah, I guess you're right, there. But if you continuously extend $\log$ to be a function from $[0,\infty]$ to $\overline{\Bbb R}$ (the extended reals) then it all works out. (I'm assuming that all of the $a_n$ are positive so that all of the summands of $\sum_{n=1}^\infty\log a_n$ are defined.) $\endgroup$ – Akiva Weinberger Feb 3 '16 at 2:37
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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 "$\ln S$"...

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