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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!

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

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

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