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.

  • $\begingroup$ Oh, that's an important point missing in my argument. Thank you! $\endgroup$
    – roman
    Mar 28 '19 at 17:13

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