- #1

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## 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....

- Thread starter ricardianequiva
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- #1

- 14

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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

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

- #2

- 334

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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.

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

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- #3

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yes. but i dont see how that helps

- #4

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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.yes. but i dont see how that helps

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

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- #5

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is there some way to actually show that the limit is 0, perhaps using definitions of limits?

- #6

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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 [tex]\epsilon > 0[/tex], you want to prove there exists an N such that for [tex]n > N[/tex],[tex]|x_n - L| < \epsilon[/tex].

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

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

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