# Cauchy sequences involving geometric series

let $\{x_n\}$ be a sequence of real numbers s.t $|x_{n+1} - x_n| \leq \alpha ^n$ $\forall n \in \mathbb{N}$ where $0 < \alpha < 1$

a) prove $|x_m - x_n | \leq \dfrac{\alpha^n}{1-\alpha}$ for $m>n$

done part a

b) Prove $|x_m - x_n | \leq \dfrac{\alpha^m + \alpha^n}{1-\alpha}$ for all $m,n$

For part b), is this the correct way to go about it?

for $m>n$, $|x_m - x_n | \leq \dfrac{\alpha^n}{1-\alpha}$ for $n>m$ $|x_m - x_n | = |x_n - x_m | \leq \dfrac{\alpha^m}{1-\alpha}$

so adding these $\forall m,n$ $|x_m - x_n| \leq \dfrac{\alpha^m + \alpha^n}{2(1-\alpha)} \leq \dfrac{\alpha^m + \alpha^n}{1-\alpha}$? Is this the correct way to do it?

c) Prove that $\{x_n\}$ is a cauchy sequence.

given epsilon, choose $N$ s.t. $\dfrac{\alpha^N + \alpha^N}{1-\alpha} < \epsilon$ hence for $m,n>N$ by the previous parts of the questions $|x_m - x_n| \leq \dfrac{\alpha^m + \alpha^n}{1-\alpha} \leq \dfrac{\alpha^N + \alpha^N}{1-\alpha} < \epsilon$ so the sequence is cauchy.

Is this correct?

• In part b), you don't have $$\lvert x_m -x_n\rvert \leqslant \frac{\alpha^m + \alpha^n}{2(1-\alpha)}$$ in general. Remove that, and it is correct. – Daniel Fischer Feb 2 '14 at 18:57
• @DanielFischer Could you explain why I have to delete it, and how you get to the final step? – Warz Feb 2 '14 at 19:50
• Say you always have $x_{n+1} = x_n + \alpha^n$. Then for $m > n$, you have $$x_m-x_n = \frac{\alpha^n - \alpha^m}{1-\alpha}.$$ For $m\to\infty$, that converges to $\frac{\alpha^n}{1-\alpha}$, and that is larger than $\frac{\alpha^m+\alpha^n}{2(1-\alpha)}$. You get the final step by noting $\max \{\alpha^m,\alpha^n\} \leqslant \alpha^m+\alpha^n$. – Daniel Fischer Feb 2 '14 at 19:57
• @DanielFischer Thank you – Warz Feb 2 '14 at 20:02

Part b: you discussed two unique different cases: for all m,n, $$|x_m - x_n| \leq \color{red} {\dfrac{\alpha^m + \alpha^n}{2(1-\alpha)}} \leq \dfrac{\alpha^m + \alpha^n}{1-\alpha}?$$ so it is correct except red part(delete it).
Part c: $$\lim _{n \to \infty}|x_{n+1} - x_n| \leq \lim _{n \to \infty}\alpha ^n=0$$ since $0 < \alpha < 1$ the result.
Uh one more note for Part B $$|x_m - x_n | \leq \dfrac{ \alpha^n}{1-\alpha}\leq \dfrac{ \alpha^n}{1-\alpha}+\color{red}{\underbrace{\dfrac{ \alpha^m}{1-\alpha}}_{\text{positive term}}}$$ again the result