What is the limit of this sequence?

[Jamin2112]

Homework Statement

I'm almost done with an arduous math problem.

Homework Equations

Not sure about any theorems I can appeal to.

The Attempt at a Solution

My sequence is basically √2, √(2+√2), √{2 + √[2+√(2+√2)]}, ...

 

[Coto]

Can you write down an equation for this sequence? Something along the lines of $x_i = f(x_{i-1})$.

 

[Jamin2112]

Can you write down an equation for this sequence? Something along the lines of $x_i = f(x_{i-1})$.

x₁ = √2, x_n+1 = √(2 + x_n).

Here's my proof so far ...

Let x₁ = √2, x_n+1 = √(2+x_n), and P(n) be the proposition that x_n < √(2+x_n).

Because 2 > 0 and √2 > 0, we know that 2 < 2 + √2. Taking the square root of both sides gives us √2 < √(2+√2), that is to say, the base case P(1).

Adding 2 to both sides of P(n) gives us xn + 2 < xn+1 + 2. Taking the square root of both sides, we then get √(x_n + 2) < √(x_n+1 + 2) = x_n+2. Thus P(n) implies P(n+1), and we can conclude that x_n < x_n+1 for all n.

Now assume x_n ≥ 2, where n > 1. Then √(2+x_n-1) ≥ 2. Squaring both sides yields 2 + x_n-1 ≥ 4, and then adding (-2) to both sides yields x_n-1 ≥ 2. Let n = 2 and our inequality implies x₁ = √2 ≥ 2, an obvious contradiction. Thus x_n < 2.

Because x_n has an upper bound, namely 2, it must have a least upper bound (see Theorem II on pg. 80). Because x_n is monotonically increasing, and because it has an upper bound, the sequence converges to its least upper bound (see Theorem III on pg. 81).

Let M be the least upper bound of x_n. We know √2 < M ≤ 2, and so 2 < M² ≤ 4. Because lim _n→∞ √(2 + x_n) = √(2 + lim _n→∞ x_n), it is true that 2 < 2 + lim _n→∞ x_n ≤ 4.

(The final paragraph is nonsense; I'm just experimenting and trying to find the limit)

 

[Coto]

Sounds like you have it figured out to me.

You can show that the limit of this sequence is 2 by showing that the sequence converges to 2.

That is,
$\forall \epsilon > 0 \ \ \ \exists N \in \mathbb{N} \ \ \ \ \mathrm{s.t.} \ \ \ \forall n \geq N, |x_n - 2| < \epsilon$.

Edit:
Hint: To do this, start by fixing epsilon, and then finding $N = N(\epsilon)$.

 

[Jamin2112]

Sounds like you have it figured out to me.

You can show that the limit of this sequence is 2 by showing that the sequence converges to 2.

That is,
$\forall \epsilon > 0 \ \ \ \exists N \in \mathbb{N} \ \ \ \ \mathrm{s.t.} \ \ \ \forall n \geq N, |x_n - 2| < \epsilon$.

Edit:
Hint: To do this, start by fixing epsilon, and then finding $N = N(\epsilon)$.

Hmmmm ... I'm still not gettin' to the limit.

Here's what I know for sure:

√2 < lim _n-->∞ x_n ≤ 2
√2 < x_n < 2 for all n ≥ 2

So if M = lim _n-->∞ x_n, then it's obvious that M - 2 ≤ 0, that is to say, x_n - M < µ for all µ > 0. I feel like I'm almost there ...

 

[Dick]

Hmmmm ... I'm still not gettin' to the limit.

Here's what I know for sure:

√2 < lim _n-->∞ x_n ≤ 2
√2 < x_n < 2 for all n ≥ 2

So if M = lim _n-->∞ x_n, then it's obvious that M - 2 ≤ 0, that is to say, x_n - M < µ for all µ > 0. I feel like I'm almost there ...

Yes, you are almost there. If x_n=sqrt(x_(n+1)+2) and you have proved x_n has a limit, M. Then doesn't that mean M=sqrt(M+2)?

 

[ZioX]

The argument is as follows
$$x_n=\sqrt{x_{n-1}+2}$$
so that if the limit does exist then
$$\lim_{n\to\infty} x_n = \lim_{n\to\infty} \sqrt{x_{n-1}+2}$$

but since sqrt(x) is continuous on R>0 we can move the limit into sqrt() on the RHS so that

$$\lim_{n\to\infty} x_n = \sqrt{\lim_{n\to\infty} x_{n-1}+2}$$

Now, since the limit on the LHS exists and say it is some value, L, then we know that the sequence of radicands (x_(n-1)+2) must converge to L+2. You should be able to see that if a sequence x_n converges to a limit L then the sequence x_(n-1) also converges to the same limit.

Therefore,

$$L= \sqrt{L+2}$$

and solve for L. This is a constraint on L and a unique value satisfying this equation must necessarily be the limit (as you have shown that it exists). Hopefully there is only one such L satisfying this equation.

This is precisely the argument that Dick is suggesting, I only elaborate so that you can see why such an argument is justified.

 

[Jamin2112]

The argument is as follows
$$x_n=\sqrt{x_{n-1}+2}$$
so that if the limit does exist then
$$\lim_{n\to\infty} x_n = \lim_{n\to\infty} \sqrt{x_{n-1}+2}$$

but since sqrt(x) is continuous on R>0 we can move the limit into sqrt() on the RHS so that

$$\lim_{n\to\infty} x_n = \sqrt{\lim_{n\to\infty} x_{n-1}+2}$$

Now, since the limit on the LHS exists and say it is some value, L, then we know that the sequence of radicands (x_(n-1)+2) must converge to L+2. You should be able to see that if a sequence x_n converges to a limit L then the sequence x_(n-1) also converges to the same limit.

Therefore,

$$L= \sqrt{L+2}$$

and solve for L. This is a constraint on L and a unique value satisfying this equation must necessarily be the limit (as you have shown that it exists). Hopefully there is only one such L satisfying this equation.

This is precisely the argument that Dick is suggesting, I only elaborate so that you can see why such an argument is justified.

This makes sense to me. You have x_n+1 = √(2+x_n), and you think, hmmmm ... It looks like our sequence will converge to 4 if x_n converges to 2. Well, what does x_n converge to? x_n = √(2+x_n-1), so it seems that x_n is converging to the same thing as x_n+1. Let M = lim _n-->∞ x_n. M = √(2+M), so M = 2.