3
$\begingroup$

Suppose that $\sum_{n=0}^\infty a_n$ is a convergent series with $a_n$ > $0$ and suppose that $(b_n){_{n\in\mathbb{N}}}$ is a bounded sequence of positive numbers. Show that $\sum_{n=0}^\infty a_n b_n$ is convergent.

Since $b_n$ is bounded and $b_n$ > $0$, can we conclude that there exists M > $0$ such that $0$ < $b_n$ < M or only that $b_n$ > $0$? If we cannot conclude that $b_n$ < M then how can we answer this question?

$\endgroup$
  • $\begingroup$ Yes you can conclude $0<b_n<M$. That is precisely what it means for a sequence of positive numbers to be bounded. $\endgroup$ – Dionel Jaime Dec 12 '17 at 22:51
  • $\begingroup$ Okay, thanks. So then we can say that $\sum a_n b_n$ < $\sum a_n M$ = M$\sum a_n$ and therefore by the comparison test $\sum a_n b_n$ is convergent? $\endgroup$ – user396611 Dec 12 '17 at 22:54
  • $\begingroup$ In fact all we need is the upper bound $M\sum a_n<+\infty$ and $\sum a_nb_n\nearrow$ which happens as $a_nb_n\ge 0$. $\endgroup$ – zwim Dec 12 '17 at 22:58
  • $\begingroup$ @DuncanRamage can we not assume the first inequality because it is stated that $a_n$ > $0$ and $b_n$ > $0$? $\endgroup$ – user396611 Dec 12 '17 at 23:04
3
$\begingroup$

Note that $$ \left|\sum_{k=n+1}^ma_kb_k\right|\leq\sum_{k=n+1}^m|a_kb_k|\leq M \sum_{k=n+1}^m|a_k|\to0 $$ as $m,n\to\infty$.

$\endgroup$
3
$\begingroup$

Since $(b_n)$ is bounded and positive, there is an $M\geq b_n >0$ for alle $n\in\mathbb{N}$.By the direct comparision test we have:

$\sum_{n=0}^\infty a_nb_n\leq \sum_{n=0}^\infty a_n\cdot M=M\underbrace{\sum_{n=0}^\infty a_n}_{<\infty}<\infty$

$\endgroup$
  • 2
    $\begingroup$ @DuncanRamage It's mentioned in the post that $b_n$ are positive. $\endgroup$ – Math Lover Dec 12 '17 at 22:59
  • $\begingroup$ I actualy do not see, why this would "prove" that non-convergent series converge. What has to be repaired? $\endgroup$ – Cornman Dec 12 '17 at 23:04
  • $\begingroup$ @Cornman I misread the question and didn't see that the an are positive. Sorry! $\endgroup$ – Duncan Ramage Dec 12 '17 at 23:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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