Suppose $(a_n)$ satisfies the condition
$$ |a_{n+1} – a_{n+2} | \leq \alpha |a_n – a_{n+1} | $$
for any $\alpha \in (0,1)$. Carefully, prove that $(a_n)$ converges.
ATTEMPT:
First, notice that $$|a_{n+1} – a_{n+2} | \leq \alpha |a_n-a_{n+1}| \leq \alpha^2 |a_{n-1} – a_n | \leq \alpha^3 |a_{n-2} – a_{n-1}| \leq … \leq \alpha^{n-1} |a_2-a_1|$$
Let $A = |a_2- a_1| \in \mathbb{R}$.
First, suppose $m>n$, then
$$ |a_n – a_m| = |a_n – a_{n+1} + a_{n+1} – a_m| = |(a_n – a_{n+1}) + (a_{n+1} – a_{n+2}) + … + (a_{m-1} – a_m)| \leq \sum_{j=1}^{m-n} \alpha^{j+n-3} \cdot A$$
Now, We know that $\sum_{j=1}^{m-n} \alpha^{j+n-3} = \alpha^{n-3} \left( \dfrac{ 1- \alpha^{m-n+1 }}{1- \alpha } \right) = \dfrac{ \alpha^{n-3} – \alpha^{m-2} }{\alpha-1} < \dfrac{\alpha^{n-3} }{\alpha-1} = \dfrac{\alpha^n}{\alpha^3(\alpha-1)}$
Thus, $|a_n – a_m| < \alpha^n \cdot \dfrac{A}{\alpha^3 (\alpha-1)} $
So, let $\epsilon > 0$ and choose $N > \dfrac{ \log(A/(\epsilon \alpha^2(\alpha-1) ) }{log(1/\alpha) } $ and for any $m>n>N$, we obtain
$$ |a_n – a_m | < \epsilon $$
So the sequence is Cauchy thus convergent.
Is this a correct solution?
Best Answer
I think your solution works, but I just wanted to give an alternative solution, since you wanted a different look at the problem. Let $y_n := x_{n + 1} - x_n$. Note that the series $\sum_{n=1}^{\infty}y_n$ is absolutely convergent, since
$$\sum_{n=1}^{\infty}|y_n| \leq y_1\sum_{n=1}^{\infty}\alpha^{n-1} < \infty,$$
using the fact that $\alpha \in (0, 1)$. Now, by telescoping, we have that
$$x_n = x_1 + \sum_{k=1}^{n - 1}y_k.$$
Therefore, taking limits of both sides, we obtain the desired result.