# The convergence of a bounded sequence ${x_n}$ satisfying $x_{n+1} - \epsilon_n \le x_n$, where $\sum_{n=1}^\infty \epsilon_n$ is absolutely convergent

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!

• Then $x_n$ converges and $\epsilon_n$'s are all non-negative. Does this lead to any contradiction? – Han Tang Apr 17 at 3:37
• We know that $|\epsilon_n|$ converges to $0$, so we must have $|x_n-x_{n+1}|<\epsilon$ for $n\geq N$. – 高田航 Apr 17 at 3:53
• The sequence must have a real $\liminf_n{x_n}$ , you can start from here. – Oolong milk tea Apr 17 at 3:54
• @高田航 $x_n-x_{n+1}$ could be a large positive number. – N. S. Apr 17 at 4:00

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.

• So $\sum_{n=1}^\infty \epsilon_n$ only needs to be convergent, not necessarily absolutely convergent? – Han Tang Apr 17 at 4:03
• @HanTang Yes, it seems so. – N. S. Apr 17 at 4:11
• Ok, it's clear now. – Han Tang Apr 17 at 4:16