# Prove that there exists a sequence $(x_n)$ such that $\sum_n a_n x_n$ diverges

So, here's a nice little result that I deduced using the closed graph theorem from functional analysis, but I'm wondering if there's a more elementary approach:

Fact: Let $(a_n)$ be a sequence with $a_n > 0$ and $a_n \to \infty$. Then there exists a sequence $(x_n)$ with $\sum |x_n| < \infty$ for which $\sum_{n=1}^\infty a_n x_n$ diverges.

I'm thinking there may be a relatively easy to construct sequence $x_n$ here, but I myself can't think of any. The reason that I know this must hold is that the map $(x_n) \mapsto (a_n x_n)$ is an unbounded operator from $\ell^1$ to $\ell^\infty$, but this provides me with no intuition as to how a suitable $(x_n)$ should be constructed.

• One can obtain an $x_n\downarrow 0$ constructively but I don't know how to improve this result. – Calvin Khor Jun 14 '16 at 15:16
• @CalvinKhor you might like \searrow, i.e. $\searrow$. And see the answer below, if you haven't already. – Omnomnomnom Jun 14 '16 at 15:20
• can't edit any more :P – Calvin Khor Jun 14 '16 at 15:22

Let $k_n$ such that $a_{k_n}\geq 2^n$.
Define $x_{k_n}= a_{k_n}^{-1}$ and $0$ elsewhere.