Find the limiting value of the sequence A sequence is given by the recurrence relation:
$$u_n = 1 + {1\over u_{n-1} +1}, u_1 = 1, n{=\ge}1$$
Work out the 2nd, 3rd and 4th term of the sequence and find the limiting value of the sequence.
So I worked out the first four terms are 1, $\frac32$, $\frac75$ and $\frac{17}{12}$ and it looks like the sequence is an irrational number 1.41421... 
Without recognising this is $\sqrt2$ just from being familiar with 1.4142... = $\sqrt2$ , is there any way  to obtain $\sqrt2$ from the recurrence relation? 
 A: Yes. Once you’ve proved that the sequence does converge, let $L$ be the limit. Then
$$L=\lim_{n\to\infty}u_n=\lim_{n\to\infty}\left(1+\frac1{u_{n-1}+1}\right)=1+\frac1{L+1}=\frac{L+2}{L+1}\;,$$
so $L(L+1)=L+2$, and $L^2=2$. It’s clear that the terms of the sequence are positive, so we must have $L=\sqrt2$.
Added: To show that the sequence converges, note that
$$u_n=1+\frac1{u_{n-1}+1}=\frac{u_{n-1}+2}{u_{n-1}+1}>u_{n-1}\;,$$
so the sequence is increasing. On the other hand, $u_1=1$, so $u_n\ge 1$ for all $n\in\Bbb Z^+$, and therefore
$$u_n=\frac{u_{n-1}+2}{u_{n-1}+1}\le\frac{u_{n-1}+2}2$$
for all $n\ge 1$. Finally, $u_1\le 2$, so by induction on $n$ we have
$$u_n\le\frac{u_{n-1}+2}2\le\frac42=2$$
for all $n\in\Bbb Z^+$. Thus, $\langle u_n:n\in\Bbb Z^+\rangle$ is a bounded increasing sequence of real numbers and therefore converges.
A: First prove it converges.  Then if it converges, we can call the limit $x$.  We then have $x = 1 + \frac{1}{x+1}$ or $x^2 = 2$ after some algebra.  The roots are $\pm\sqrt{2}$.  Argue that it's the positive one.
