# Show that sequence is Cauchy

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.

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.

• Thanks! To find the limit of this sequence, I tried to plug in n = m+1 in the Cauchy sequence definition. Is this the right approach? – Einstein the troll Feb 8 at 23:17
• No. That is not something you can do. The Cauchy sequence definition has a "for all" condition - restricting like that gets you an inconclusive result "this might be a Cauchy sequence, or it might not". – jmerry Feb 8 at 23:24

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,$$a_n=\frac{3-\left(\frac{1}{3}\right)^{n-1}}{2}$$.

• I think you mean subtract off the equation. – IAmNoOne Feb 8 at 23:54
• No,you must add them,look carefully at the signs. – Alexdanut Feb 8 at 23:55

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

If $$n then $$|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}}$$