Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Below is a question that I came across while studying real analysis. Given that $\{a_k\}_{k=1}^{\infty}$ is a sequence of complex numbers where $\sum_k{a_k}{b_k}$ converges for every complex sequence $\{b_k\}_{k=1}^{\infty}$ such that $\lim_kb_k=0$. Prove that $\sum_k|{a_k}|<\infty$. As a matter of fact as I am very new to this field. I need some help. Thanks.

share|improve this question

marked as duplicate by mixedmath, Gerry Myerson, t.b., Henning Makholm, Asaf Karagila Nov 13 '11 at 5:34

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
And interestingly enough, my answer to that question is the same as Gerry's below. –  mixedmath Nov 13 '11 at 3:09
1  
@mixedmath, great minds think alike. I should have looked for a duplication. –  Gerry Myerson Nov 13 '11 at 3:40
add comment

1 Answer 1

up vote 1 down vote accepted

Suppose $\sum|a_k|$ diverges. Choose $M_1\lt M_2\lt\dots$ such that $\sum_{M_i\le k\lt M_{i+1}}|a_k|\ge1$. Choose $b_k$ such that $a_kb_k\ge0$ and $b_k=1/i$ for $M_i\le k\lt M_{i+1}$. Then $b_k\to0$ but $\sum_{M_i\le k\lt M_{i+1}}a_kb_k\ge1/i$, so $\sum a_kb_k$ diverges.

share|improve this answer
    
Thanks for the answer. –  smanoos Nov 13 '11 at 3:34
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.