2
$\begingroup$

Let {$a_k$} and {$b_k$} be sequences. Suppose that $|b_{k+1}-b_k| \leq a_k$ for all $k \in \mathbb{N}$ and that $\sum_{n=1}^{\infty} a_k$ converges. Prove that {$b_k$} converges.

I know I need to prove by showing that {$b_k$} is Cauchy, but I'm not sure if I can use {$a_k$} is Cauchy since it converges (because it only says the sum of {$a_k$} converges).

Can anyone show me a formal proof of this? Thanks!

$\endgroup$
2
$\begingroup$

Let $\epsilon > 0$. By definition, there exists $n,m \in \mathbb{N}$ such that $\sum_{k=n}^m a_k < \epsilon$,(Since $a_k$ is nonnegative) so $$ |b_m - b_n| = \left|\sum_{k=n}^{m-1} (b_{k+1} - b_k)\right| \leq \sum_{k=n}^{m-1} |b_{k+1} - b_k| \leq \sum_{k=n}^{m-1} a_k < \epsilon $$

$\endgroup$
  • $\begingroup$ are you trying to show that $\sum_{k=0}^n b_k$ converges? Because I mean to show that the sequence {$b_k$} converges, aren't they two different things? $\endgroup$ – mathmathmath Mar 25 '16 at 2:58
  • $\begingroup$ You need to consider that $b_k$ may be negative don't you? $\endgroup$ – Matematleta Mar 25 '16 at 2:58
  • $\begingroup$ @mathmathmath If the sum converges then the sequence $\to 0$. $\endgroup$ – Henricus V. Mar 25 '16 at 2:59
  • $\begingroup$ @Chilango Then repeat the proof for $-b_k$. $\endgroup$ – Henricus V. Mar 25 '16 at 3:00
  • $\begingroup$ @HenryW. I am required to prove by showing {$b_k$} is Cauchy, can you show me that way? $\endgroup$ – mathmathmath Mar 25 '16 at 3:00

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.