$\sum a_n$ converges, $a_n \in \mathbb{R}$, then there exists real sequence $b_n$ such that $b_n\rightarrow +\infty$ and $\sum a_n b_n$ converges.

Question

Assume that $\sum a_n$ converges, $a_n \in \mathbb{R}$, then there exists real sequence $b_n$ such that $b_n\rightarrow +\infty$ and $\sum a_n b_n$ converges.

Same to be easy at first thought, can we find such $b_n$ represented by $a_n$?

Answer 1

Credit: Rudin's "Principles of Mathematical Analysis", Chapter 3, Exercise 12.

Let $$r_n=\sum^\infty_{m=n}a_n.$$ Then $\sum\frac{a_n}{\sqrt{r_n}}$ is convergent.

Answer 2

This answer is intended to furnish an intuition to @Quang Hoang's answer.

1. First consider an integral analogue of the question.

If $f > 0$ and $\int_{0}^{\infty} f(x) \, \mathrm{d}x < \infty$, then can we find a function $g \geq 0$ such that $g(x) \to \infty$ and $\int_{0}^{\infty} f(x)g(x) \, \mathrm{d}x < \infty$?

A useful trick in this direction is to play with antiderivatives of $f$. For example, let $F(x) = \int_{x}^{\infty} f(t) \, \mathrm{d}t$. Since the integral converges, $F$ is well-defined as a real-valued function and satisfies $F' = -f$. Then

$$ \sqrt{F(0)} = \int_{0}^{\infty} \left( -\frac{\mathrm{d}}{\mathrm{d}t}\sqrt{F(t)} \right) \, \mathrm{d}t = \int_{0}^{\infty} \frac{f(t)}{2\sqrt{F(t)}} \, \mathrm{d}t. $$

So by letting $g(x) = \frac{1}{2\sqrt{F(x)}}$, it follows that $g(x) \to \infty$ as $x \to \infty $ yet the integral of $fg$ over $[0, \infty)$ converges.

2. Now the construction in @Quang Hoang's answer is precisely the series analogue of the above heuristics. Indeed, suppose $a_n > 0$ and $\sum_{n=1}^{\infty} a_n < \infty$. Let

$$ r_n = \sum_{k=n}^{\infty} a_k $$

and note that

$$ \sqrt{r_n} - \sqrt{r_{n+1}} = \frac{a_n}{\sqrt{r_n} + \sqrt{r_{n+1}}} \geq \frac{a_n}{2\sqrt{r_n}}. $$

(This is the series analogue of $\sqrt{F(x)} - \sqrt{F(x+h)} = \int_{x}^{x+h} \frac{f(t)}{2\sqrt{F(t)}} \, \mathrm{d}t$.) This allows to choose a $(b_n)$ that satisfies the desired properties.

Answer 3

HINT: Because $\sum a_n$ converges to $A$, we can partition it into sets

$$\{a_1, \dots, a_{N_1}\}, \{a_{N_1+1}, \dots, a_{N_2}\}, \dots$$

where $$\sum_{i={N_k+1}}^{N_{k+1}} a_i \approx \frac{A}{4^k}.$$

Then if you pick $b_n$ as the sequence $B_k$ but with a lot of repetition, you can make $\sum a_n b_n \approx \sum \frac{A}{4^k} B_k$.

What remains to be done is choosing the numbers $N_k$ precisely and making the approximate equality more precise by changing it to an inequality and/or modifying $A/4^k$. There might be a snag with alternating sums, that is sums that have negative terms, but I think the essence of the idea is here.