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Consider the sequence so defined by induction:

$x_0 = 1$

$x_{n+1} = \sqrt{1+x_n}$

Prove that it is bounded and find the limit towards +infinity.

To me it seems to approach phi but don't know how to prove it.

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    Before find the limit you also need to prove that the sequence is incresing. Just bounded is not enough. – 2017-01-10

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Boundedness is an easy induction. Note that $x_0 = 1 < 2$. Suppose $0 < x_n < 2$. Then $x_n < \sqrt{1 + x_n} < \sqrt{3} < 2$. It follows that $x_n < 2$ for all $n$, so by the monotone convergence theorem $x_n$ converges. As for the limit, note that if the sequence $a_n = f^n(x)$ converges to $L$, where $f$ is continuous on an interval containing $L$ and $f^n$ is the $n$th composition of $f$ with itself, we must have $f(L) = L$. What can you conclude about the limit from this?

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    Yes yes the boundness somehow I had a rough idea. My problem is the limit. Really can't get it, what can I conclude? – 2017-01-10
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    Solve the equation $\sqrt{1 + L} = L$. – 2017-01-10
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    Can I even if I am in a limit? – 2017-01-10
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    I mean, do I have to justify more? – 2017-01-10
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    I just told you that the limit is some solution to that equation. – 2017-01-10
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See that the function defined in $\Bbb R_{+}^{*}$

$$f(x)=\sqrt{1+x} \rightarrow f'(x)=\frac{1}{2\sqrt{1+x}}>0$$

So $f(x)$ is monotonic increasing and then the sequence $x_{n+1}=\sqrt{1+x_n}$ is also monotonic increasing.

Once we know that $x_{n+1}>x_n$ then

$$x_{n+1}^2>x_n^2 \rightarrow x^2_n-x_n-1<0 \rightarrow \frac{1-\sqrt{5}}{2}

So the sequence is bounded and increasing then it converge to $L$.

Doing $n \rightarrow \infty$ at $x_{n+1}^2=x_n+1$ we get:

$$L^2=L+1 \rightarrow L=\frac{1\pm\sqrt{5}}{2}$$

But $L>0$ then:

$$L=\frac{1+\sqrt{5}}{2}$$