# Recursive sequences convergence

1. Mar 1, 2008

### ricardianequiva

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. The attempt at a solution

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

2. Mar 1, 2008

### 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.

Last edited: Mar 1, 2008
3. Mar 1, 2008

### ricardianequiva

yes. but i dont see how that helps

4. Mar 1, 2008

### jhicks

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?

Last edited: Mar 1, 2008
5. Mar 1, 2008

### 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?

6. Mar 1, 2008

### 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.

Last edited: Mar 1, 2008