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

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?

• $n=0$ doesn't make sense. – zhw. Apr 25 '20 at 16:28
• @zhw. Sorry, the question has been edited now. – Shaurya Apr 25 '20 at 16:30
• Did you try the comparison test ? – Learning Apr 25 '20 at 16:34
• @Learning I couldn't find a better series to bound the summand which could possibly help in determining the convergence. – Shaurya Apr 25 '20 at 16:43

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 $$\sum_{n \in \mathbb{N}} \frac{e^{na_n}}{n^2} \geq \sum_{n \in S} \frac{e^{na_n}}{n^2} = \sum_{n \in \mathbb{N}} \frac{e^{n^{2/3}}}{n^4} = + \infty.$$

If $$\sum_{n=1} a_n$$ converges and $$(a_n)$$ is decreasing, we must have that $$\lim_{n \rightarrow \infty} na_n = 0$$ (see this answer). Choose $$N \in \mathbb{N}_0$$ such that for all $$n \geq N: a_n \leq \frac{1}{n}$$. We get $$\sum_{n =1}^\infty \frac{e^{n a_n}}{n^2} = \sum_{n =1}^{N-1} \frac{e^{n a_n}}{n^2} + \sum_{n=N}^\infty \frac{e^{n a_n}}{n^2} \leq C + \sum_{n=N}^\infty \frac{e^{1}}{n^2} < \infty$$ When $$\sum_{n=1}^\infty a_n$$ diverges, the series $$\sum_{n =1}^\infty \frac{e^{n a_n}}{n^2}$$ can still converge, take for exampe $$a_n = 1/n$$.

• The proof for $na_{n}\underset{n\to +\infty}{\longrightarrow}0$ assumes that $\left(a_{n}\right)$ is decreasing. – CHAMSI Apr 25 '20 at 17:31
• Yes, I dind't look at the required assumptions. – abcdef Apr 25 '20 at 21:19

Let us first try it using comparison test. Suppose that $$u_n=\frac{e^{n a_n}}{n^2}$$ and $$v_n=\frac{1}{n^2}$$.

Now, $$\lim\limits_{n\to\infty} \frac{u_n}{v_n}=\lim\limits_{n\to\infty}\frac{\frac{e^{n a_n}}{n^2}}{\frac{1}{n^2}}=\lim\limits_{n\to\infty}e^{n a_n} = \lim\limits_{n\to\infty} \{1+\frac{n a_n}{1!}+\frac{(na_n)^2}{2!}+\cdots\}$$

1. If we have $$\{a_n\}$$ to be a decreasing sequence of positive real numbers and $$\sum_{n=1}^{\infty}a_n$$ converges, then $$\lim_{n \rightarrow \infty} n a_n = 0$$. This gives $$\lim\limits_{n\to\infty} \frac{u_n}{v_n}=1$$, so $$\sum u_n$$ and $$\sum v_n$$ will have same nature.
2. Let $$\sum a_n$$ is divergent. Take $$a_n=\frac{\log n}{n}$$, we get $$\sum \frac{e^{na_n}}{n^2}=\sum \frac{1}{n}$$.
3. See Raoul's answer when $$\sum a_n$$ is convergent.

We conclude that the series $$\sum_{n=1}^{\infty}\frac{e^{na_n}}{n^2}$$ may or may not be convergent.