Neat series convergence result

Question

This is a nice problem on series convergence that I recently stumbled upon. Given a non-negative sequence of real numbers $(a_n)$ such that $$\sum_{n=1}^\infty a_n < \infty,$$ show that there exists a non-decreasing sequence of non-negative numbers $b_n$ such that $$b_n \to \infty \quad\text{ and } \quad \sum_{n=1}^\infty a_n b_n < \infty.$$ In other words, for every convergent series with non-negative terms, there is another convergent series with "substantially larger" terms.

I have a solution (see below), but maybe someone else has a different, simpler, and/or more elegant solution.

Answer 1

By convergence of the series there exists a strictly increasing sequence of indices $(n_k)$ such that $$\sum_{n=n_k}^\infty a_n < 2^{-k}.$$ Then $$\sum_{k=1}^\infty \sum_{n=n_k}^\infty a_n < \sum_{k=1}^\infty 2^{-k} = 1.$$ Since series with non-negative terms can be reordered arbitrarily, we get $$1 > \sum_{n=1}^\infty \sum_{k\in I_n} a_n = \sum_{n=1}^\infty a_n b_n,$$ where $I_n = \{ k\in\mathbb{N}: n_k \le n \}$ and $b_n = \#I_n$ is the number of elements in $I_n$. We have that $n_k$ is strictly increasing and $n_k \to \infty$ as $k\to\infty$, so $b_n$ is non-decreasing and $b_n \to \infty$.

Answer 2

Here is another way to do it. Let $$r_n = \sum_{k=n}^\infty a_k.$$ Then $(r_n)$ is a non-increasing sequence with $r_n \to 0$. We can assume $r_n >0$ for all $n$ (otherwise the claim is trivially true), and observe that $$\sqrt{r_n} - \sqrt{r_{n+1}} = \frac{r_n-r_{n+1}}{\sqrt{r_n} + \sqrt{r_{n+1}}} = \frac{a_n}{\sqrt{r_n} + \sqrt{r_{n+1}}}.$$ Define $$ b_n = \frac1{\sqrt{r_n} + \sqrt{r_{n+1}}}.$$ Then $(b_n)$ is a non-decreasing, positive sequence with $b_n \to \infty$, and $$\sum_{k=1}^n a_k b_k = \sum_{k=1}^n (\sqrt{r_k} - \sqrt{r_{k+1}})=\sqrt{r_1} - \sqrt{r_{n+1}},$$ so $$\sum_{k=1}^\infty a_k b_k = \sqrt{r_1} < \infty$$