I'm given the following problem: Does the series $\sum\limits_{n=1}^\infty a_n$ converge if $a_n \geq 0$ and $\sqrt[n]{a_n}<1$ for $n \in \mathbb N$?
(It's a bit unclear to me if "for $n \in \mathbb N$" means for all $n$, or for $n$ after some $N$, I'm assuming the former since it's a stronger condition)
My question: Is this solution correct?
My attempt:
The series can both diverge and converge, because of the following 2 examples
Example for divergence: consider a diverging series $\sum\limits_{n=1}^\infty \frac{1}{n+1}$, then $a_n = \frac{1}{n+1} \geq 0$ and $\sqrt[n]{a_n}=\sqrt[n]{\frac{1}{n+1}}<1$ for all $n \in \mathbb N$.
Example for convergence: consider $\sum\limits_{n=1}^\infty \Bigl(\frac{1}{n^2}\Bigr)^n$, which converges by the root test: $\lim\limits_{n \to \infty}\sqrt[n]{\Bigl(\frac{1}{n^2}\Bigr)^n}=0<1$. and $a_n=\Bigl(\frac{1}{n^2}\Bigr)^n\geq0$ and $\sqrt[n]{a_n}=\sqrt[n]{\Bigl(\frac{1}{n^2}\Bigr)^n}<1$ for all $n \in \mathbb N$.
