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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.

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    It is probably easiest to attack this in contraposed form: If $\sum |a_k|=\infty$, then construct a sequence $\phi_k$ such that $\sum |a_k|\phi_k =\infty$, yet $\phi_k\to 0$.2011-11-05

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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 $a_j$.

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    Good (+1). Perhaps a bit clearer: consider the sequence $\{n_k\in\mathbb{R}\}$ so that \displaystyle\sum_{n_k\le n. Let $\phi_n=\frac{1}{k}\operatorname{signum}(a_n)$ for n_k\le n.2011-11-05