Statement: If a bounded sequence $\{x_n\}_{n=0}^\infty$ in $\mathbb{R}$ satisfies $x_{n+1} - \epsilon_n \le x_n$ for $n \in \mathbb{N}$, where $\sum_{n=1}^\infty \epsilon_n$ is an absolute convergent series, then $\{x_n\}_{n=0}^\infty$ is convergent.
I think the statement is true and there are two cases:
(1) There are finitely many positive $\epsilon_n$'s.
Then $\exists N\in \mathbb{N}$ s.t. $x_{n+1} - x_n \le \epsilon_n \lt 0 \; \forall n \ge N$. So $\{x_n\}_{n=0}^\infty$ is decreasing when $n \ge N$. Since it is bounded, it converges.
(2) There are infinitely many positive $\epsilon_n$'s.
This is where I am stuck. I think since $\sum_{n=1}^\infty \epsilon_n$ is absolutely convergent, $\epsilon_n$ converges to $0$. But how to proceed?
Any hint would be appreciated!
Let $S_n= \epsilon_1+\epsilon_2+...+\epsilon_{n-1}$. Then, $S_n$ is convergent, and hence also bounded.
Define $y_n=x_n-S_n$. Then, $y_n$ is the difference of two bounded sequences and hence bounded. Moreover,
$$y_{n+1}-y_n=x_{n+1}-x_n -(S_{n+1}-S_n)=x_{n+1}-x_n-\epsilon_n \leq 0$$
This shows that $y_n$ is monotonic.
Therefore, $y_n$ is monotonic and bounded, and hence convergent.
It follows that $x_n=y_n+S_n$ is the sum of two convergent sequences, thus convergent.