Recursive sequences convergence

[ricardianequiva]

Homework Statement

Let the sequence {a_n} defined by:

a_n+1 = a_n/[sqrt(0.5a_n + 1) + 1]

Prove that {a_n} converges to 0

Homework Equations

The Attempt at a Solution

I tried manipulating the equation but to no avail...

 

[jhicks]

If the sequence approaches 0, then lim(n->infinity) a_n should be 0 as should a_n+1 correct?

If the series converges, then a_n and a_n+1 should both approach the limit L of the sequence as n becomes large.

 

[ricardianequiva]

yes. but i don't see how that helps

 

[jhicks]

yes. but i don't see how that helps

Ok let me try it a different way. Suppose instead of explicitly labeling the sequence as a sequence, I say that the sequence is actually a function f(x) which takes x_n and transforms it into x_n+1.

For this sequence, f(x)=\frac{x}{\sqrt{x/2+1}+1}. If the sequence converges to a limit L, then clearly f(L)=L. Does that make more sense?

 

[ricardianequiva]

yes, i can see how that shows that the limit is 0, but i don't think its rigorous enough.
is there some way to actually show that the limit is 0, perhaps using definitions of limits?

 

[jhicks]

I had to delete my previous post because I found an error in it, but I think the argument can be made by comparison. Using the definition of a limit, for every \epsilon > 0, you want to prove there exists an N such that for n > N,|x_n - L| < \epsilon.

The question is vague enough and seems to ignore the lack of convergence if x_n = -2 so I'm going to restrict my argument to x_n > 0. Clearly for arbitrary w and u > 0 such that w > u, then \frac{1}{\sqrt{0.5w+1}+1} < \frac{1}{\sqrt{0.5u+1}+1}. Using this, you can place an upper bound on the ratio of subsequent terms of the sequence x_{n+1} = \frac{x_n}{\sqrt{0.5x_n+1}+1} to say that x_{n+1} \leq \frac{1}{2} x_n, which is much easier to work with.

I don't think attempts to directly apply the definition of a limit without using this kind of workaround would be met with much success.