# 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.

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!

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 $$|b_m - b_n| = \left|\sum_{k=n}^{m-1} (b_{k+1} - b_k)\right| \leq \sum_{k=n}^{m-1} |b_{k+1} - b_k| \leq \sum_{k=n}^{m-1} a_k < \epsilon$$
• 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? Mar 25 '16 at 2:58
• You need to consider that $b_k$ may be negative don't you? Mar 25 '16 at 2:58
• @mathmathmath If the sum converges then the sequence $\to 0$. Mar 25 '16 at 2:59
• @Chilango Then repeat the proof for $-b_k$. Mar 25 '16 at 3:00
• @HenryW. I am required to prove by showing {$b_k$} is Cauchy, can you show me that way? Mar 25 '16 at 3:00