Show that sequence is Cauchy

Question

I need to show that the above sequence is a Cauchy sequence. However, when using the definition of a Cauchy sequence, I get that $s(n) - s(m)$ is equal to some complicated sigmal notation expression, which I need to show to be less than $\epsilon$ for every $\epsilon>0$ for $n,m > N$. Any help would be appreciated.

Answer 1

We don't need an exact value of $a_n-a_m$; we just need an estimate. Assuming WLOG that $n>m$, we get $a_n-a_m=(a_{m+1}-a_m)+(a_{m+2}-a_{m+1})+\cdots+(a_n-a_{n-1}) = \frac1{3^m}+\frac1{3^{m+1}}+\cdots+\frac1{3^{n-1}}$.

For fixed $m$, how big can that get? Well, clearly, the worst case comes as $n\to\infty$. What is the infinite sum $\frac1{3^m}+\frac1{3^{m+1}}+\cdots+\frac1{3^n}+\cdots$?

This series (a geometric series) is one of the ones you should know exactly. In most cases, you'll want to compare the series to something - but here, the geometric series is one of the standard things to compare to.

Answer 2

Alternatively, you can find a formula for $a_n$ and this makes your problem easier.Let's write the recurrence relation for $a_n$,$a_{n-1}$,...,$a_2$ :
$a_n=a_{n-1}+\frac{1}{3^{n-1}}$
$a_{n-1}=a_{n-2}+\frac{1}{3^{n-2}}$
.................................
$a_2=a_1+\frac{1}{3}$
After we add the above lines we get that $a_n=1+\frac{1}{3}+...\frac{1}{3^{n-1}}$,which is a geometric progression.Hence,$a_n=\frac{3-\left(\frac{1}{3}\right)^{n-1}}{2}$.

Answer 3

Hint That complicated sigma notation expression is a geometric sum and it is easy to calculate.

If $n <m$ then $$|s(m)-s(n)|=\sum_{k=n}^{m-1}\frac{1}{3^k}=\frac{1}{3^n} \sum_{k=n}^{m-1}\frac{1}{3^{k-n}}=\frac{1}{3^n} \sum_{j=0}^{m-n-1}\frac{1}{3^{j}}=\frac{1}{3^n} \frac{1-\frac{1}{3^{m-n}}}{1-\frac{1}{3}}$$