How do I show convergence here?

Question

I have a sequence $(x_n)$ and some $X_n$, where:

$X_n:=|x_1-x_2|+|x_2-x_3|+\ldots +|x_n-x_{n+1}|\leq M$

where $M>0$.

How do I show that $X_n$ converges? I have some idea that since adding up all the differences of terms in the sequence is less than or equal to some value, I think the sums get less and less... If they got bigger, then they wouldn't fit under some constant $M$ (and would diverge). I'm just not sure how to write this in math...

Also, I have to show that $x_n$ converges, but how do I get this from the $X_n$ result?

Answer 1

Monotone Convergence Theorem. Monotone bounded sequences converge (Wikipedia page here).

We see that $X_n$ is monotone (non-decreasing), since $X_{n+1}-X_n=|x_{n+1}-x_n|\geq 0$. We also got $0\leq X_n\leq M$, and thus, $X_n$ converges.

Answer 2

For last question: If $X_n$ converges then $|x_n-x_{n+1}|$ converges to $0$ and $\{x_i\}$ is a Couchy sequence. If you consider field of real numbers then $x_n$ converges

Answer 3

Let $m := sup~\{X_n : n \in N\} \leq M$.

Let $\epsilon > 0$.

Then $c := m - \epsilon < m$. As $m$ is the smallest upper bound and $c < m$, $c$ cannot be an upper bound. So there exists $N_{0} \in \mathbb{N}: m \geq X_n > c \Rightarrow |m - X_{N_0}| = m - X_{N_0} < \epsilon$ and as $X_n$ is increasing $\forall n \geq \mathbb{N}: X_{N_0} \leq X_n \Rightarrow |X_n - m| = m - X_n \leq m - X_{N_0} < \epsilon$.

So $\forall \epsilon > 0~\exists N \in \mathbb{N}~\forall n \geq N: |X_n - m| < \epsilon~\Leftrightarrow \lim_{n \rightarrow \infty} X_n = m \leq M$.

Second part:

Because $X_n$ converges, $(X_n)_{n \in \mathbb{N}}$ has to be cauchy $\Leftrightarrow~\forall \epsilon > 0~\exists N \in \mathbb{N}~\forall n,m\geq N: |X_n - X_m| < \epsilon$.

Now let $\epsilon > 0$ then there exists $N_{0} \in \mathbb{N}~\forall n,m \geq N_{0}: |X_n-X_m| < \epsilon$. Let $i,k \geq N_{0}+1$ with $i > k$. Then by the definition of $X_n$, telescoping sum and triangle inequality: $|x_i-x_k|=|\sum_{l=k+1}^{i} x_{l-1} - x_{l}| \leq \sum_{l=k+1}^{i} |x_{l-1} - x_{l}| = X_{i-1} - X_{k-1} = |X_{i-1} - X_{k-1}| \leq \epsilon$. So $(x_n)_{n \in \mathbb{N}}$ is Cauchy (check the definition above for $x_n$) and therefore converges.