Suppose a sequence $a_{n}$ has a property: there exist constant C and K, with $0<K<1$,

Question

The book I am using for my Advance Calculus course is Introduction to Analysis by Arthur Mattuck.

Suppose a sequence $a_{n}$ has a property: there exist constant C and K, with $0<K<1$, such that

$|a_n - a_{n+1}| < CK^{n}$, for n>>1.

Prove that $a_{n}$ is a Cauchy sequence.

This is my rough proof to this question. I was wondering if anybody can look over it and see if I made a mistake or if there is a simpler way of doing this problem. I want to thank you ahead of time it is greatly appreciated.So lets begin.

Proof:

Answer 1

It's almost correct. First you need to add "Given $\epsilon > 0$" at the beginning. This seems nitpicky, but the point is that if someone gives you any small number, you can show for sufficiently large $n, m, |a_n-a_m|$.

Next, you have a correct proof that $|a_n-a_{n+p}|\leq CK^n\frac{1-K^{p+1}}{1-K}$ (nice job this is the crux of the proof!!). What you need to do still is show that as $n\to \infty$, the value on the right, $CK^n\frac{1-K^{p+1}}{1-K}$ is less than that given fixed $\epsilon$. Since $0\leq K<1$ we know $K^n \to 0$. So you need to state that for all large $p, \frac{1-K^{p+1}}{1-K}$ stays bounded.

All the stuff under the "For, ..." at the end can be omitted. Make sense?

Answer 2

A sequence $(a_{n})$ of reals is called Cauchy iff for every $\varepsilon > 0$ there is some $N \geq 1$ such that $|a_{n} - a_{m}| < \varepsilon$ for all $n,m \geq N$. Moreover, in this question it is required that $C > 0$.

Your argument may be simplified as follows: Let $\varepsilon > 0$; if $n$ is sufficiently large (say $\geq$ some $N_{1} \geq 1$) and if $p \geq 1$, then by assumptions we have $$ |a_{n} - a_{n+p}| \leq CK^{n}(1+K + \cdots + K^{p}) \leq CK^{n}(\frac{1}{1-K}), $$ which is $< \varepsilon$ if in addition we have $$ n > N_{2} := \bigg| \frac{\log \frac{\varepsilon (1-K)}{C}}{\log K} \bigg|; $$ therefore, taking $N := \max \{ N_{1}, N_{2}+1 \}$ suffices.