$\{x_n\}$ 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!

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    $\begingroup$ 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. $\endgroup$ – Thorgott Mar 21 at 16:54
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    $\begingroup$ @Thorgott that’s an important point you brought up. Thank you. I will update the post once I get back to my laptop $\endgroup$ – roman Mar 21 at 16:58

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.


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