let $b_n$ be a positive sequence. s.t. for each $\epsilon>0$ there is N s.t. for each $m>n>N$:
$ \sum_{k=n}^mb_k <\epsilon$
and let $a_n$ be a sequence that fulfills that for each $n\in \mathbb N$:
$|a_{n+1}-a_n|\le b_n$
prove that $a_n$ converges.
my solution: by looking at the first terms in the sequence $a_n$ i noticed that :
$$ |a_{n+1}-a_n| \le b_n$$ $$.$$ $$.$$ $$ |a_{m-1}-a_{m-2}| \le b_{m-2}$$ $$ |a_{m}-a_{m-1}| \le b_{m-1}$$ $$ |a_{m+1}-a_{m}| \le b_{m}$$
using the first given information we get that by choosing every desired $\epsilon$ there is an $N$ s.t. for every $m>n>N$ we get:
$|b_n+b_{n+1} + b_{n+2} + .. + b_m|<\epsilon$ (because every element in the sum is positive)
by summing the inequalities I wrote above:
$|a_{m+1}-a_{n}| = |a_{n+1} - a_n +...+ a_{m-1} - a_{m-2} + a_m - a_{m-1} + a_{m+1} - a_m| \le |a_{n+1}-a_n| + |a_{n}-a_{n-1}|+...+|a_{m-1}-a_{m-2}| + |a_{m}-a_{m-1}| +|a_{m+1}-a_{m}| \le |b_n+b_{n+1} + b_{n+2} + .. + b_m|<\epsilon$
then we define $m+1 = n+p$, where $p \in \mathbb Z$, and we can do this because $m>n$. so by Cauchy's test, $a_n$ converges.
do you think my solution is okay? any kind of help would be appreciated.
$$|a_{k}-a_h| = |a_k - a_{k+1} + a_{k+1} - \dots +a_{h+1} - a_h| \leq $$ $$ \leq |a_k - a_{k-1}| + |a_{k-1} - a_{k-2}| + \dots +|a_{h+1} - a_h| \leq b_{k-1} + \dots + b_h = \sum_{i=h}^{k-1} b_i \leq \varepsilon $$ for $k,h$ large enough. So $a_n$ is Cauchy, therefore converges.