Okay, so I'm stuck on this question which is part (b) to the question and was wondering if someone could help me out with it, I have solved the first part as seen below, (would you also be able to check (a); I feel my communication is lacking as the proof is very small), thanks!
Question
Consider a sequence $(b_n)_{n=1}^{\infty}$ of non-zero real numbers. By definition, the infinite product $\prod_{n=1}^{\infty}$ converges if the sequence $(p_n)_{n=1}^{\infty}$, where
$p_n=\prod_{k=1}^{n}b_k$
converges to some non-zero number.
(a) Prove that the convergence of $\prod_{n=1}^{\infty}b_n$ implies $\lim_{n\to\infty}b_n=1$
Working (a)
If $p_n \rightarrow p\neq 0$,
then $b_n=\frac{p_n}{p_{n-1}}=\frac{\prod_{k=1}^{n}b_k}{\prod_{k=1}^{n-1}b_k}\rightarrow\frac{p}{p}=1$
QED
(b) Assume $b_n>0$ for all $n\in\mathbb{N}$. Prove that $\prod_{n=1}^{\infty}b_n$ converge if and only if
$\sum_{n=1}^{\infty}\ln (b_n)$
converges. (You may use without proof the fact that $\ln(x)$ is a continuous function on $(0, \infty)$.)
Thanks again :)
Use the fact that $\log (xy)=\log x+\log y$. For you question, you have $$\log\left(\lim_{N\to \infty} \prod_{n=1}^N b_n\right)=\lim_{N\to \infty}\log\prod_{n=1}^N b_n=\lim_{N\to\infty}\sum_{n=1}^N \log b_n$$ and you can interchange the limit and the logarithm because $\log$ is a continuous function on $(0,\infty)$. Here, you use your hypotheses on $b_n$. Then, the convergence of one side is dependant on the convergence of the other and vice versa.
I hope this helps.