Why does this sequence converge to 0?

Question

So I have a convergent sequence in $(x_n)_n$ in $\mathbb{R}$. We define $y_n= x_{n+1}-x_n$. I have to show $(y_n)_n$ is convergent.

Now I understand that the proof uses the triangle inequality $|x_{n+1}-x_n| \leq |x_n- x| + |x - x_{n+1}| < 2\epsilon$ since $(x_n)_n$ converges to $x$.

So this means that $x_{n+1}$ will converge to $x_n$. If it converges to $x_n$ and $x_n$ converges to $0$, then can we assume that $(y_n)_n$ will converge to $0$?

Now, the professor used in the last part of the triangle inequality the $\leq 2 \epsilon$. Why do we not say it is strictly smaller than $\epsilon$?

Answer 1

Suppose $x_n \to x.$ We want to show that $y_n$ is convergent. Indeed, we have

$$ |x_{n+1} - x_n| = |x_{n+1} - x + x - x_n| \leq |x_{n+1} - x| + |x_n - x| $$

by the old "add zero" trick. Now recall that the inequality $|x_n - x| \leq \frac{\epsilon}{2} $ when $n \geq N$ (for some natural number $N$). Now for large enough $n$ we have

$$ |x_{n+1} - x_n| + |x_n - x| \leq \epsilon $$

as needed. While no inference in the limit of $y_n$ is made, we can guess that its limit is zero and manipulate the inequalities. The guess comes from the limit laws (the limit of the difference is the difference of limits, provided both exist). The sequence $x_{n+1}$ is just a shift of $x_n$ by one term, and the limits are the same.

(Note that I can replace the $\frac{\epsilon}{2}$ by $\epsilon$ - we just want the difference to be small enough. As a professor I once had says, "It's the same amongst friends.")