# Hard calculus proofs!

1. Feb 24, 2010

### _Steve_

I've got two calculus proofs that I can't seem to get! I was wondering if you guys could help me out a bit...

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

Suppose $$x_{n}$$ is the sequence defined recursively by

$$x_{1}=0$$ and $$x_{n+1}=\sqrt{5 + 2x_{n}}$$ for $$n=0, 1, 2, 3, ....$$

Prove that $$x_{n}$$ converges and find its limit

The attempt at a solution

So far, by taking the limit of both sides "$$l$$", I've found the limit

$$l=\sqrt{5 + 2l}$$
$$l=\frac{-2 +\sqrt{24}}{2}$$

but I still have to prove that the sequence converges. I've tried induction hypothesis using

$$x_{n} \leq x_{n+1} \leq \frac{-2 +\sqrt{24}}{2}$$

and then I attempt to modify $$x_{n}$$ and $$x_{n+1}$$ such that the equation becomes (in the end):

$$x_{n+1} \leq x_{n+2} \leq \frac{-2 +\sqrt{24}}{2}$$

but I can never get it to work out!

2.
The problem statement, all variables and given/known data

The second proof I'm having trouble with is basically one of the same type...

Let $$x_{n}$$ be the sequence of real numbers defined recursively by

$$x_{0} = 0$$ and $$x_{n+1} = \ln (2 + x_{n})$$ for $$n=0, 1, 2, 3,.....$$

Show that $$x_{n} \leq x_{n+1} \leq 2$$

The attempt at a solution

Again, I try to modify it such that i get
$$\ln (2 + x_{n}) \leq \ln (2 + x_{n+1}) \leq 2$$

but I keep ending up with
$$\ln (2 + x_{n}) \leq \ln (2 + x_{n+1}) \leq \ln (2 + 2)$$

Is there a trick to calculus proofs?

Thanks!
Steve

2. Feb 25, 2010

### netheril96

In my memory,to prove such thing with induction,you need an induction hypothesis that xn is not only bounded above,but also bounded below

Add a proper lower bound,and try mathematical induction again

3. Feb 25, 2010

### Staff: Mentor

You have a sign error in your calculation of L.
Starting with
$$L=\sqrt{5 + 2L}$$
$$\Rightarrow L^2=5 + 2L$$
$$\Rightarrow L^2 - 2L - 5 = 0$$

By the Quadratic Formula, this gives
$$L = \frac{2 \pm \sqrt{4 - (-20)}}{2} = 1 + \sqrt{6} \approx 3.4495$$
The other solution, 1 - sqrt(6), is negative, and can be discarded. The recursive relation won't give negative values.

Since $$L = lim_{n \to \infty} x_{n + 1}$$
this shows that the sequence converges, and gives its limit.

4. Feb 25, 2010

### _Steve_

I figured them out! Thanks Mark!

Is a lower bound really necessary? I assumed not, because it will obviously be larger than $$x_{1}=0$$ since the sequence is increasing

5. Feb 25, 2010

### VeeEight

It is clear that it is bounded below. You can also consider monotone convergence theorem (although, with Mark's remarks, the proof is correct as it is).

6. Feb 25, 2010

### LCKurtz

I don't think so. What it shows is that if the sequence converges, then that value of L is its limit. But you must still argue that it converges. For consider x1 = 1 and xn+1 = 2xn. That argument applied to this would give L = 2L which would imply the limit exists and is 0.

7. Feb 25, 2010

For 1: you already know $$x_0 = 0 < x_1 = \sqrt{5 + 2 x_0} = \sqrt{5}$$
Now use induction: assume $$x_k < x_{k+1}$$ and show $$x_{k+1} < x_{k+2}$$. (Hint: look at simplifying $$x_{k+2}^2 - x_{k+1}^2$$
For 2: try to show something like $$x_n < 5$$ for all n.