$\{x_n\}$ 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!
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.