$\sum_{k=1}^{\infty}a_kb_k$ converges for every bounded sequence {$b_k$}, prove that $\sum_{k=1}^{\infty}a_k$ converges absolutely.

Let $\sum_{k=1}^{\infty}a_k$ be a series of real numbers. Suppose that $\sum_{k=1}^{\infty}a_kb_k$ converges for every bounded sequence {$b_k$} ? Prove that $\sum_{k=1}^{\infty}a_k$ converges absolutely.

Here is my thought process:

Since {$b_k$} is bounded, every |$b_k$| $\leq$ M for all k.

So $\sum_{k=1}^{\infty}|a_kb_k|$ = $\sum_{k=1}^{\infty}|a_k||b_k|$ $\leq$ $\sum_{k=1}^{\infty}|a_k|M$ = $M$$\sum_{k=1}^{\infty}|a_k$| and since $\sum_{k=1}^{\infty}|a_kb_k|$ converges, $\sum_{k=1}^{\infty}|a_k|$ also converges.

However, I am having a hard time figuring out how to prove absolute convergence. I realize that they are similar questions on this website but they're asking to prove that $\sum_{k=1}^{\infty}a_kb_k$ converges, rather than $\sum_{k=1}^{\infty}a_k$ converging absolutely. Would really love any help, I'm quite stuck!

• Why'd you drop the absolute value bars...? – Simply Beautiful Art Mar 5 '17 at 22:35
• Have you tried looking at a specific sequence $(b_k)_k$, depending on $(a_k)_k$, such that (1) $(b_k)_k$ is bounded, and (2) $a_kb_k = \lvert a_k\rvert$ for every $k$? – Clement C. Mar 5 '17 at 22:36

Hint: define $b_k=\frac{|a_k|}{a_k}$ if $a_k\neq 0$, and $b_k=0$ if $a_k=0$.
• The sequence $\{b_k\}$ in my answer is bounded, and $a_kb_k=|a_k|$. – carmichael561 Mar 5 '17 at 23:26