How to approach this proof on convergence of infinite product? 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. 
 A: 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.
