Convergence of $\sum_{n=0}^{\infty}\frac{e^{na_n}}{n^2}$

Question

Question: Suppose you are given a sequence $\displaystyle (a_n)_{n=1}^{\infty}$ such that $a_n\gt 0$ $\forall n\in\Bbb{N}$. Suppose we also know that $\displaystyle \lim_{n\to\infty}a_n=0$. Now if we are given that $\displaystyle \sum_{n=1}^{\infty}a_n$ converges then what can we say about the convergence of $$\sum_{n=1}^{\infty}\frac{e^{na_n}}{n^2}?$$ What if $\displaystyle \sum_{n=1}^{\infty} a_n$ diverges?

I tried to check the convergence using ratio test but it wasn't much helpful. Moreover the root test was also inconclusive. Can anybody drop some hints please?

Answer 1

If $\sum a_n$ diverges, it clearly does not work: just take $a_n = 1/\sqrt{n}$.

If $\sum a_n$ converges, it still does not work, without extra assumptions like $(a_n)$ being decreasing (see other answer). For instance, consider $(a_n)$ defined by $a_n = 1/n^{2/3}$ if $n$ is a perfect square, and $a_n = 1/n^2$ otherwise. Denote $S$ the set of perfect squares. Then we have the following.

• The series $\sum a_n$ converges since $$ \sum_{n \in \mathbb{N}} a_n = \sum_{n \in S} a_n + \sum_{n \notin S} a_n \leq \sum_{n \in \mathbb{N}} \frac{1}{n^2} + \sum_{n \in \mathbb{N}} \frac{1}{n^{4/3}} < + \infty. $$
• On the other hand $$ \sum_{n \in \mathbb{N}} \frac{e^{na_n}}{n^2} \geq \sum_{n \in S} \frac{e^{na_n}}{n^2} = \sum_{n \in \mathbb{N}} \frac{e^{n^{2/3}}}{n^4} = + \infty. $$