this was given as an exercise: Prove that if $\sum_{n=1}^{\infty} |a_n|$ converges and $(b_n)^{\infty}_{n=1}$ is a bounded sequence, then $\sum_{n=1}^{\infty} |a_nb_n|$ converges

This is what i was thinking:

since $\displaystyle\sum a_n = \lim_{k\to\infty}\sum_{n<k} a_n$, and when multiplying by a constant it can jump into the limit.

then If $b_n\le c$ then $\sum a_nb_n\le \sum ca_n\le c\sum a_n$, and then $\sum_{n=1}^{\infty} |a_nb_n|$ converges.

  • 2
    $\begingroup$ You are essentially correct, though I would put absolute values: $$\sum_{n=1}^\infty |a_n b_n| \leq \sum_{n=1}^\infty C|a_n| = C \sum_{n=1}^\infty |a_n| < \infty.$$ $\endgroup$ – snar Mar 10 '14 at 3:28
  • $\begingroup$ The sequence $S_N= \sum_1^N|a_nb_n|$ increases and is bounded, hence convergent. $\endgroup$ – AD. Mar 9 '15 at 11:09

Here is another way using the Cauchy's criterion:

Let $M>0$ such that $|b_n|< M$. Given $\varepsilon>0$, choose $n_0$ such that

$$\sum_{n=p+1}^q|a_n|< \varepsilon/M $$ for all $q<p<n_0$


$$\bigg|\sum_{n=p+1}^qa_nb_n\bigg|\le\sum_{n=p+1}^q|a_nb_n|\le M\sum_{n=p+1}^q|a_n|< \varepsilon$$


Absolute values are critical: consider $a_n=(-1)^n/n$ and $b_n=(-1)^n$. Luckily you are given absolute convergence so use it to show $\sum a_nb_n$ converges absolutely.


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