Proving Convergence with Cauchy Sequence Method

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

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

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

You can use the given property on each of those and use induction to show the general case.
 
HallsofIvy said:
To show that a sequence is Cauchy you must show that |a_m- a_n| goes to 0 as m and n go to infinity independently (in particular, you cannot assume that m= n+1).

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

You can use the given property on each of those and use induction to show the general case.
asaaaa
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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