# $\{x_n\}$ is a bounded above sequence such that $x_{n+1} - x_n \ge a_n$, where $\sum a_k$ converges. Prove $x_n$ converges.

$$\{x_n\}$$ is a bounded above sequence satisfying the following property: $$x_{n+1} - x_n \ge \alpha_n\tag1$$ where $$\alpha_n$$ is such that $$\exists \lim_{n\to\infty} \sum_{k=1}^n \alpha_k$$ Prove $$\{x_n\}$$ converges.

I'm trying to generalize the idea from this question. Below are some thoughts.

First denote: $$S_n = \sum_{k=1}^n \alpha_k$$

Since $$S_n$$ is convergent then it must be bounded both below and above. Let: $$y_n = x_n - S_{n-1}$$

Since $$x_n$$ is bounded above and $$-S_n$$ is also bounded above (by convergence of $$S_n$$), then it must follow that $$y_n$$ is also bounded above: $$\exists M\in\Bbb R: y_n \le M, \forall n\in\Bbb N \tag2$$ Rewrite $$(1)$$ as: $$x_{n+1} \ge x_n + \alpha_n$$

Now subtract $$S_n$$ from both sides: $$\underbrace{x_{n+1} - S_n}_{y_{n+1}} \ge x_n - S_n + \alpha_n = \underbrace{x_n - S_{n-1}}_{y_n}$$

That means $$y_n$$ is monotonically increasing. By $$(2)$$ we know $$y_n$$ is bounded. Finally by monotone convergence theorem: $$\exists \lim_{n\to\infty}y_n \implies \exists\lim_{n\to\infty}(x_n - S_{n-1})$$

Which in terms means that $$x_n$$ is also convergent. I would like to ask for a verification of the proof above or/and point to mistakes in case of any. Thank you!

• You claim $y_n=x_n-S_{n+1}$ is bounded above. A sufficient condition is that both $x_n$ and $-S_{n-1}$ are bounded above, i.e. that $x_n$ is bounded above and that $S_{n-1}$ is bounded below. Your argument would not work if $S_n$ were only bounded above, but the convergence of $\sum\alpha_k$ ensures boundedness both ways, so the argument can easily be fixed. – Thorgott Mar 21 at 16:54
• @Thorgott that’s an important point you brought up. Thank you. I will update the post once I get back to my laptop – roman Mar 21 at 16:58

## 1 Answer

Your proof is correct, nicely done!

Just as an illsutration of how $$\limsup$$ and $$\liminf$$ can shorten such arguments:

As $$x_n\geq x_m+\sum_{k=m}^n \alpha_k$$ for all $$m\leq n$$ we have $$\liminf_{n\to\infty}x_n\geq \liminf_{n\to\infty}\left(x_m+\sum_{k=m}^n \alpha_k\right)=x_m+\sum_{k=m}^\infty \alpha_k$$ and then taking $$\limsup_{m\to\infty}$$ on the right side we get $$\liminf_{n\to\infty}x_n\geq\limsup_{m\to\infty}\left(x_m+\sum_{k=m}^\infty \alpha_k \right)=\limsup_{m\to\infty}x_m+\lim_{m\to\infty}\sum_{k=m}^\infty \alpha_k=\limsup_{m\to\infty}x_m$$ so $$\{x_n\}$$ converges.