Question: For what values of $p >0$ does the infinite product $\prod_{n=1}^\infty \frac{1}{n^p}$ converge? For what of $p >0$ does the infinite product converge absolutely.
My attempt: Note that if $a_n \geq0$ for all $n$ then $$\prod_n (1-a_n)\ \text{converges} \iff \sum_n a_n\ \text{converges}$$ Writing $n^{-p} = 1 - (1 - n^{-p})$ where $a_n = 1 - n^{-p}$ is nonnegative for each $n\in \mathbb{N}, p > 0$ then as $\lim_{n\to \infty} a_n = \lim_{n\to \infty} 1 - \frac{1}{n^p} = 1$, by divergence test, $\sum_n a_n = \sum_n (1 - n^{-p})$ does not converge. So, there does not exists any such $p > 0$ in which $\prod_p \frac{1}{n^p}$ converges or converge absolutely.
Could I check if there are any gaps in my proof, as this was listed as an exam problem, yet my solution seems quite trivial.