Convergent Sequence

• Aug 11th 2009, 07:18 AM
roshx
Convergent Sequence
Hi, I think I am close to the answer for this but I am missing something:

Suppose $x_{n}$ is a sequence satisfying $|x_{n+1}-x_{n}|<\frac{1}{2^n}$ for all n. Show that $x_{n}$ is a convergent sequence.

So to prove it is convergent, I need to prove it is a Cauchy sequence, ie for all $\epsilon$ there is an N such that $|x_{m}-x_{n}|<\epsilon$ for all $n,m>N$.

Now $|x_{m}-x_{n}|= |x_{m}-x_{m-1}+x_{m-1}-...+x_{n+1}-x_{n}|$ and the triangle inequality gives $|x_{m}-x_{n}|\leq|x_{m}-x_{m-1}|+|x_{m-1}-x_{m-2}|+...+|x_{n+1}-x_{n}|$.
Then using the inequality given in the question I get $|x_{m}-x_{n}|<\frac{1}{2^{m-1}}+\frac{1}{2^{m-2}}+...+\frac{1}{2^n}<\frac{1}{2^n}+\frac{1}{2^n}+ ...+\frac{1}{2^n}=\frac{m-n}{2^n}$.
This is where I get stuck, I think I need to get rid of that m-n somehow.
• Aug 11th 2009, 07:31 AM
Opalg
Yes, you are quite close to the answer, right up to the inequality $|x_{m}-x_{n}|< \frac{1}{2^{m-1}}+\frac{1}{2^{m-2}}+...+\frac{1}{2^n}$ (where m>n). Here, instead of estimating each individual term (replacing each term by the largest term $1/2^n$), write the terms in the reverse order $\frac{1}{2^n} + \frac{1}{2^{n+1}} + \frac{1}{2^{n+2}} + \ldots + \frac{1}{2^{m-1}}$. This is a (finite) geometric series, whose sum is less than the infinite sum $\sum_{r=n}^\infty \frac1{2^r} = \frac1{2^{n-1}}$.
• Aug 11th 2009, 07:42 AM
roshx
Ah cheers, I never think of using geometric series!