Proving Convergence with Cauchy Sequence Method

[JamesF]

Hello all. I'm having trouble on the following homework problem. It seems like it should be easy, but I'm just now sure how to approach it

Homework Statement

Let [tex](s_n)[/tex] be a sequence st [tex]|s_{n+1} - s_n | < 2^{-n}, \forall n \in \mathbb{N}[/tex]

show that [tex](s_n)[/tex] converges

The Attempt at a Solution

well I thought the easiest way to prove it would be to show it's a Cauchy sequence and therefore convergent, but perhaps that's the wrong approach. What's a good starting point for a problem like this?

 

[Dick]

A good starting point is to show it's Cauchy. But you already knew that. Just do it. Cauchy means |s_n-s_m|<epsilon for n and m greater than N. The maximum difference between s_n and s_m involves summing a lot of large powers of 1/2. Use the triangle inequality.

 

[HallsofIvy]

To show that a sequence is Cauchy you must show that [itex]|a_m- a_n|[/itex] goes to 0 as m and n go to infinity independently (in particular, you cannot assume that m= n+1).

But [itex]|a_{n+2}- a_n|\le |a_{n+2}- a{n+1}|+ |a{n+1}- a_n|[/itex], [itex]|a_{n+3}- a_n|\le |a_{n+3}- a_{n+2}|+ |a_{n+2}- a{n+1}|+ |a{n+1}- a_n|[/itex], etc.

You can use the given property on each of those and use induction to show the general case.

 

[HallsofIvy]

To show that a sequence is Cauchy you must show that [itex]|a_m- a_n|[/itex] goes to 0 as m and n go to infinity independently (in particular, you cannot assume that m= n+1).

But
[tex]|a_{n+2}- a_n|\le |a_{n+2}- a{n+1}|+ |a{n+1}- a_n|[/tex]
[tex]|a_{n+3}- a_n|\le |a_{n+3}- a_{n+2}|+ |a_{n+2}- a{n+1}|+ |a{n+1}- a_n|[/tex], etc.

You can use the given property on each of those and use induction to show the general case.

asaaaa