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I have the following question I was given on a tutorial sheet to revise Infinite Series, and on it Infinite Products are introduced, as is the following question:

Assume $\,b_{n}>0\,$ for all $\,n \in \mathbb{N}$ . Prove that $\prod_{n=1}^{\infty} b_{n}$ converges if and only if $$\sum_{n=1}^{\infty} \ln b_{n}$$ converges.

I was wondering if anyone could give me some pointers on how to start this; as it's an if and only if proof, obviously I need to consider the case where the infinite product $b_n$ converges but $log b_n$ doesn't, and vice versa, but I'm having an incredibly difficult time wrapping my head around product series.

I know from a class that $\log(x)$ is continuous on $(0, \infty)$, which I think might help in showing the forward implication, but other than that I'm completely lost.

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  • $\begingroup$ For any partial sum, you have $$\ln \prod_{k \le N} b_k = \sum_{k \le N} \ln b_k$$ simply by the property of $\ln$ $$\ln xy = \ln x + \ln y$$ $\endgroup$ – Crostul May 22 '17 at 11:28
  • $\begingroup$ hint: $\prod_{n=1}^kb_n=exp(\sum_{n=1}^kln(b_n))$, and $exp()$ is continuous, hence if one partial sum converges, then... $\endgroup$ – Evargalo May 22 '17 at 11:30
  • $\begingroup$ @Evargalo But watch out for the case where $\lim_{k\to \infty}\prod_{n=1}^k b_n$ exists whereas $\prod_{n=1}^\infty b_n$ does not! $\endgroup$ – Hagen von Eitzen May 22 '17 at 11:33
  • $\begingroup$ @Crostul Alright, so for the reverse implication, if the $\prod b_n$ converges, then surely $\log\prod b_n$ must as well correct? $\endgroup$ – Pixel Rain May 22 '17 at 11:43
  • $\begingroup$ @HagenvonEitzen : how exactly do you define the convergence of $\prod_{n=1}^\infty b_n$ ? $\endgroup$ – Evargalo May 22 '17 at 12:05
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The forward implication is false without the additional requirement that the limit $\,\prod_{k=1}^\infty b_k\,$ is strictly positive. Take for example $\,b_n = 1/2 \gt 0\,$, then $\,\prod_{n=1}^\infty b_n = 0\,$ but $\,\sum_{n=1}^\infty \ln b_n \to -\infty\,$.

With that caveat, let $\,p_n = \prod_{k=1}^n b_k\,$ and $\,s_n=\sum_{k=1}^n \ln b_k\,$, then what can be proved is that $\,p_n\,$ converges to a non-zero limit as $\,n \to \infty\,$ if and only if $\,s_n\,$ converges. To that end, note that:

  • $\;\displaystyle s_n=\ln p_n\,$ by the property of the logarithm that $\,\ln ab = \ln a + \ln b\,$;

  • $\;\displaystyle p_n=e^{s_n}\,$ since the natural logarithm and the exponential are inverses of each other.

$\log(x)$ is continuous on $(0, \infty)$, which I think might help in showing the forward implication

Intuition is right. Suppose that $\,p_n\,$ converges to $\,p \gt 0\,$ i.e. $\,\lim_{n \to \infty} p_n = p\,$, then $\,\lim_{n \to \infty} s_n\,$ $\,= \lim_{n \to \infty} \ln p_n\,$ $\, = \ln \big(\lim_{n \to \infty} p_n\big) = \ln p\,$ by continuity of $\ln$ on $\mathbb{R}^+$, so $\,s_n\,$ is convergent.

The reverse implication works much the same. Suppose that $\,\lim_{n \to \infty} s_n = s\,$, then $\,\lim_{n \to \infty} p_n\,$ $\,= \lim_{n \to \infty} e^{s_n}\,$ $\, = e^{\lim_{n \to \infty} s_n} = e^s\,$ by continuity of the exponential function, so $\,p_n\,$ is convergent.

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  • $\begingroup$ An infinite product is called convergent if the limit is positive... $\endgroup$ – N. S. Jun 16 '19 at 13:22
  • $\begingroup$ @N.S. You must mean "non-zero" rather than "positive". Even that depends on the definitions you use, There were times and places where an infinite product was considered convergent iff the limit of partial products existed, which is what my first paragraph was about. $\endgroup$ – dxiv Jun 17 '19 at 3:44

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