# 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!

• Sequence is monotone, what sign is $x_n-x_{n-1}$ for all $n$? It's the same. So $\sigma_n = |x_{n+1}-x_1|$ – Jakobian Mar 26 '19 at 18:21

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