real analysis - Prove that convergence of a sequence implies boundedness of its variation.

A sequence $\{x_n\}$ is said to have bounded variation if a sequence $\sigma_n$ is bounded, where $\sigma_n$ is defined as:

$$\sigma_n = |x_2 - x_1| + |x_3 - x_2| + \cdots + |x_{n+1} - x_n|,\ n\in\Bbb N$$
Prove that any monotone bounded sequence $\{x_n\}$ has bounded variation.

I've started with taking a look at $\sigma_n$, since it is a sum of absolute values then it must follow that:

$$\sigma_{n+1} \ge \sigma_n$$
Thus $\sigma_n$ is monotonically increasing. To show a sequence is bounded it is sufficient to show that it converges.

Here is what we wan't to show eventually:

$$\exists\lim_{n\to\infty}x_n = L_1 \implies \exists M\in\Bbb R: \sigma_n \le M\ \forall n\in\Bbb N$$
It is given that $\{x_n\}$ is bounded and monotonic, thus it converges by Monotone Convergence Theorem. Define a new sequence:
$$y_n = x_{n+1} - x_n$$
By convergence of $x_n$ it follows that:
$$\exists \lim_{n\to\infty}y_n = \lim_{n\to\infty}(x_{n+1} - x_n) = 0$$
But then it also follows that $y_n$ converges absolutely:
$$\lim_{n\to\infty}|y_n| = \lim_{n\to\infty}|x_{n+1} - x_n| = 0$$
Let's now fix some number $p \in \Bbb N$ and consider the following expression:

$$\sigma_{n+p} - \sigma_n = \sum_{n+1}^{n+p} |y_k|$$
Now consider the limit of RHS:
$$\lim_{n\to\infty}\sum_{n+1}^{n+p}|y_k| = 0$$

Then if follows:

$$\lim_{n\to\infty}|\sigma_{n+p} - \sigma_n| = 0$$

Thus $\sigma_n$ satisfies Cauchy's Criteria hence convergent, hence bounded.

I would like to ask for a verification of the proof above, and point to mistakes in case of any, or suggest a solution in case the above makes no sense. Thank you!

Answer

If $\{x_n\}$ is increasing,

$$\sigma _n=x_{n+1}-x_1.$$
Since $\{x_n\}$ is bounded, then of course $\sigma _n$ is bounded...