Does This Sequence Converge Given the Conditions?

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Matt B.
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Homework Statement

: [/B]Prove that if xn is a sequence such that |xn - xn+1| ≤ (1/3n), for all n = 1,2,..., then it converges.

Homework Equations

: [/B]The definition of convergence.

The Attempt at a Solution

:[/B] I attempted to prove this by induction, so I am clearly far off the mark here. Any advice would be appreciated.
 
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Our definition of convergence provided is the following:

{an} converges to A if ∀ε>0, ∃a positive integer N such that n ≥ N => | an - A | < ε.

I'm not sure how I am supposed to use this in the example above?
 
Matt B. said:
Our definition of convergence provided is the following:

{an} converges to A if ∀ε>0, ∃a positive integer N such that n ≥ N => | an - A | < ε.

I'm not sure how I am supposed to use this in the example above?

Have you ever heard of Cauchy sequences?
 
Ray Vickson said:
Have you ever heard of Cauchy sequences?
Yes.
 
Ray Vickson said:
OK... so?

Do I just assume that xn can be substituted for the usual xn and that xn+1 be substituted for xm and attempt to find the n,n+1 ≥ N such that | xn - xn+1 | < ε, given ε > 0?
 
Matt B. said:
Do I just assume that xn can be substituted for the usual xn and that xn+1 be substituted for xm and attempt to find the n,n+1 ≥ N such that | xn - xn+1 | < ε, given ε > 0?

No, that is not what a Cauchy sequence is all about. A sequence is Cauchy if, given any ##\epsilon > 0## there is an ##N = N(\epsilon)## such that ##|x_n - x_m|< \epsilon## for all ##n,m > N##. Note that this is ##|x_n - x_m|##, not just ##|x_n - x_{n+1}|##.

You might guess that if I mention Cauchy sequences, that must be for a good reason.