# Prove a sequence with bounded variation converges.

A sequence is said to have bounded variation if: $$\exists M \in\Bbb R: \sigma_n = |x_2 - x_1| + |x_3 - x_2| + \cdots + |x_{n+1} - x_n| \le M,\ \forall n\in\Bbb N$$ Prove that boundedness of variation implies convergence of $$\{x_n\}$$

This question is based on my previous question here, where I needed to prove 'convergence implies boundedness of variation'. Now I want to do the opposite.

First, note that $$\sigma_n \ge 0,\ \forall n\in \Bbb N$$. The sequence is also convergent by monotone convergence theorem, because $$\sigma_n$$ is monotonically increasing: $$\sigma_n \le M,\ \sigma_{n+1} \ge \sigma_n \implies \exists \lim_{n\to\infty}\sigma_n = L$$

Then $$\sigma_n$$ satisfy Cauchy's criteria, thus we may fix any $$p \in\Bbb N$$, such that: $$\lim_{n\to\infty}(\sigma_{n+p} - \sigma_n) = 0$$

Consider the difference: $$\sigma_{n+p} - \sigma_n = \sum_{k=n+1}^{n+p}|x_k - x_{k-1}|$$

Writing the limit for both sides: $$\lim_{n\to\infty}(\sigma_{n+p} - \sigma_n) = \lim_{n\to\infty}\sum_{k=n+1}^{n+p}|x_k - x_{k-1}| = 0$$

And that is only possible in case every term is the sum tends to 0 no matter what $$p$$ we choose, which means: $$\exists \lim_{n\to\infty} |x_{n+p} - x_{n}| = 0$$

Therefore $$x_n$$ is Cauchy, hence convergent.

I would like to ask for verification of my proof. If the above is invalid, what would be a proper proof?

Your argument is fine until "And that is only possible in case every term in the sum tends to 0...". There is a missing step (probably it was already clear for you, but you just forgot to write). Namely, by the triangle inequality we have $$\vert x_{n+p} - x_n \vert \leq \sum_{k=n+1}^{n+p} \vert x_k - x_{k-1} \vert =\sigma_{n+p} - \sigma_n.$$ Then you can conclude that $$(x_n)_{n\in \mathbb{N}}$$ is Cauchy and thus convergent.