Convergence of an infinite product using logarithm

Question

Suppose we have $(a_n)_{n\geq 1}$ is a positive real sequence, and we know that $\sum\log a_n$ converges. How can we conclude that the product $$ \prod_{n\geq 1}\sqrt{\frac{2}{a_n+a_n^{-1}}} $$ converges? My approach was to let $b_n :=\log\frac{2}{a_n+a_n^{-1}}$ and the product becomes $$ \prod_{n\geq 1}\sqrt{\frac{2}{a_n+a_n^{-1}}}=\exp\left(\frac{1}{2}\sum_{n\geq 1} b_n\right). $$ But, I am stuck at concluding $\sum b_n$ converges. May I know how should I proceed? Thank you!

Answer 1

From $$t_n:=\log a_n\to0,$$ we derive $$u_n:=\frac{a_n+a_n^{-1}}2-1=(\cosh t_n)-1\sim\frac{t_n^2}2,$$ and $$-b_n=\log(1+u_n)\sim u_n.$$

Therefore, if $\sum|t_n|$ (hence also $\sum t_n^2$) converges, so does $\sum b_n.$

However, if $\sum t_n$ is only semi-convergent, then $\sum t_n^2$ (hence also $\sum b_n$) may diverge. Example: $t_n=\frac{(-1)^n}{\sqrt n}.$