# Convergence of $\sum{a_kb_k}$ if $\sum{a_k}$ converges and $\sum{b_k}$ absolutely converges.

Convergence of $\sum{a_kb_k}$ if $\sum{a_k}$ converges and $\sum{b_k}$ absolutely converges.

I tried to think that Since $\sum |b_k|$ is bounded

I thought that $\sum a_k b_k$ $<$ $S\sum a_k$. Is it a correct inequality?

If both sequences are just conditionally converge, then do $\sum{a_k b_k}$ converge?

Because $\sum_ka_k$ converges, we have $\lim_ka_k=0$ so that $|a_k|$ is bounded by some $M>0$. If you assume that $\sum_k|b_k|$ converges, then $\sum_k|b_k|$ is Cauchy. Then, because $$\left|\sum_{k=m}^na_kb_k\right|\leq \sum_{k=m}^n|a_k||b_k|\leq M \sum_{k=m}^n|b_k|,$$ we infer that $\sum_k a_kb_k$ is also Cauchy and thus converges.
You need the absolute convergence of $\sum_k b_k$. André's answer provides a neat example.
• I appreciate to your answer. then by inequality above, would $\sum a_k b_k$ converge absolutely? – Luke Lee Sep 21 '14 at 15:31
• I think you are right: $|\sum_m^n |a_kb_k||=\sum_m^n|a_k||b_k|$ so the argument above still applies. – Kim Jong Un Sep 21 '14 at 15:37
Conditional convergence is not enough. Let $a_n=\dfrac{(-1)^n}{\sqrt{n}}$ and $b_n=\dfrac{(-1)^{n+1}}{\sqrt{n}}$.