Is it possible that $\sum n a_n^2$ converges but $\sum a_n$ diverges?

Question

Yesterday I was thinking about a problem, when an interesting question appeared:

Does there exist a sequence $a_n \geq 0$ of non-negative real numbers such that $\sum_{n \geq 1} n a_n^2 < \infty$ and $\sum_{n \geq 1} a_n = \infty$?

After thinking about it for a while, I couldn't come up with either an example of this or a proof that this is impossible. Is this known (I didn't manage to find it online)?

Comments: One thing I noticed is that, if $p > 1$ is fixed, we may assume without loss of generality that $$\frac{1}{n^p} <a_n < \frac{1}{n} \forall n.$$ The reason the following:

Assume there exists $a_n$ that satisfies the above two conditions, and let $S = \left\{n \in \mathbb{N}, n \geq 1 | a_n < \frac{1}{n}\right\}$. Then for every $n \geq 1, n \notin S$, we have $a_n \leq na_n^2$, thus $$\sum_{n \notin S} a_n \leq \sum_{n \geq 1} n a_n^2 < \infty $$

Therefore, since $\sum_{n \geq 1} a_n = \infty$, it follows that $\sum_{n \in S} a_n = \infty$. If we take $b_n = \begin{cases} a_n, &n \in S \\ 0, &n \notin S \end{cases}$, $b_n$ is a non-negative sequence also satisfying the above two properties. Therefore, we may assume wlog that $a_n < \frac{1}{n} \forall n$.

Now, if $p>1$ is fixed we may also assume wlog that $a_n > \frac{1}{n^p}$, because if $a_n$ satisfies the two properties, then $$b_n = \begin{cases} a_n, &a_n > \frac{1}{n^p} \\ \frac{1}{n^{(p+1)/2}}, &a_n \leq \frac{1}{n^p} \end{cases}$$ also satisfies them.

Answer 1

You could probably have success with $a_n=\frac1{(n+1)\log(n+1)}$.

$$\sum_{n\ge 2}\frac1{n\log(n)^2}$$

converges by twice applying the Cauchy condensation test or the integral test.

Answer 2

Let $$a_n=\frac1{n\ln n}$$ then using the integral test the series $$\sum_n a_n$$ is divergent but the series $$\sum_n na_n^2$$ is convergent.

Answer 3

An example such that $\sum\limits_nna_n^2$ converges and $\sum\limits_na_n$ diverges: $a_n=\frac1{n\log n}$.