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!

  • $\begingroup$ Sequence is monotone, what sign is $x_n-x_{n-1}$ for all $n$? It's the same. So $\sigma_n = |x_{n+1}-x_1|$ $\endgroup$
    – Jakobian
    Mar 26, 2019 at 18:21

1 Answer 1


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


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