If $\sum c_n$ converges absolutely and $k_n$ is bounded, does $\sum c_nk_n$ converge absolutely?

Suppose a series $\sum\limits c_n$ converges absolutely and a sequence $k_n$ is bounded. Will the sequence $\sum\limits c_nk_n$ converge absolutely?

Since $k_n$ is bounded there must exist an integer $M>0$ such that $\vert k_n \vert \leq M$. We then get $\vert \sum\limits c_n k_n \vert \leq \sum\limits \vert c_nk_n\vert \leq M\sum\limits \vert c_n \vert < M\epsilon$. Does this show that $\sum\limits\vert c_nk_n\vert$ converges absolutely?

• i think i should say $\lvert \sum\limits \vert c_nk_n\vert$ $\lvert$ also Sep 28 '13 at 1:15
• Abel criterion does the trick. Sep 28 '13 at 20:32
• Please don't deface your own question. Sep 30 '13 at 1:04
• Apr 7 '14 at 12:10

3 Answers

Yes, your proof is entirely correct, though I think you mean $c_n$ everywhere you have $a_n$. In fact, this is a special case of an inequality known as Holder's Inequality, with $p = 1$ and $q = \infty$.

Another proof uses Cauchy's criterion: A series $\sum_j c_j$ converges if and only if, for any $\epsilon > 0$, there is a $N(\epsilon)$ such that $|\sum_{j=m}^n c_j| < \epsilon$ for all $N(\epsilon) < m < n$. Since $|k_j| < M$, $|\sum_{j=m}^n k_j c_j| < M\epsilon$ for $N(\epsilon) < m < n$.

I would say this is enough, but if you want get the bound to be $\epsilon$ instead of $M \epsilon$, replace $\epsilon$ by $\epsilon/M$ to get $|\sum_{j=m}^n k_j c_j| < \epsilon$ for $N(\epsilon/M) < m < n$.

In $\vert \sum\limits c_n k_n \vert \leq \sum\limits \vert c_nk_n\vert \leq M\sum\limits \vert c_n \vert < M\epsilon$ I guess you mean the sums for $n=M,M+1,\dots$, the Cauchy Property.