How to prove $\sum\limits_{k=1}^{\infty}|\alpha_{k}|\lt \infty$, given that $\sum \limits_{k=1}^{\infty}\alpha_{k}\phi_{k}$ converges ...?

I am solving some problems in a text, I come across this question. I thought I will not take much of my time on it, but that is not the case.

Question:

Prove that if $\sum \limits_{k=1}^{\infty}\alpha_{k}\phi_{k}$ is convergent whenever $\lim_{k\to \infty}\phi_{k}=0$, then $\sum \limits_{k=1}^{\infty}|\alpha_{k}|\lt \infty$

I am sorry for this question, it seems too simple but I do not know how to tackle it. Please I need just a hint for this.

Solution 1:

This is a great question!. Here's a brief sketch.

Assume that $\sum |a_k| = \infty$. Consider $\phi_k$ s.t. if $\sum_{n_k}^{n_{k+1}} |a_j| < k$, then $\phi_k (n) = \frac{1}{k}$ for $n_k < n < n_{k+1}$. Btw, $k \in \mathbb{N}$ throughout.

What happens to $\sum a_j \phi_k$?

I note in addition that the sign of the $\phi_k$ should match their respective $a_j$.