Limit of the sequence defined by a recurrence Given a recurrence formula for an arithmetic sequence, $$U_{n} = \frac{1}{2+U_{n-1}}$$
Show that$$\frac{1}{2+\frac{1}{2+\frac{1}{2+\frac{1}{2+ ...}}}} = (SomeGivenValue)$$
How can we solve questions like this?

The method is to make a substitution $U_n=\frac{T_n}{T_{n+1}}$ and you would get much more tangible
$$T_{n+1}^2-2T_nT_{n+1}-T_n^2=0$$
$$(T_{n+1}-T_n-\sqrt{2}T_n)(T_{n+1}-T_n+\sqrt{2}T_n)=0$$
$$T_{n+1}-T_n-\sqrt{2}T_n=0$$
$$T_{n+1}-T_n+\sqrt{2}T_n=0$$
These two you solve classically assuming $T_n=a^n$ and when you substitute and solve you have that $a_1=1-\sqrt{2}$ and $a_2=1+\sqrt{2}$ which gives
$$T_n=c(1-\sqrt{2})^n+d(1+\sqrt{2})^n$$
and the solution follows. Set initial conditions and you have the solution.
$$U_n=\frac{c(1-\sqrt{2})^n+d(1+\sqrt{2})^n}{c(1-\sqrt{2})^{n+1}+d(1+\sqrt{2})^{n+1}}$$
Now all you need to do to get to the answer of some value is to find:
$$\lim\limits_{n \to \infty}\frac{c(1-\sqrt{2})^n+d(1+\sqrt{2})^n}{c(1-\sqrt{2})^{n+1}+d(1+\sqrt{2})^{n+1}} = \sqrt{2}-1$$
Since $c$ and $d$ depend on the initial value (in your case that would be $U_0 = \frac{1}{2}$) this proves that the initial value is irrelevant.

Three years later....

A small technical glitch
We cannot have
$$ \frac{c}{d}=\left ( \frac{\sqrt{2}+1}{\sqrt{2}-1} \right )^n= \left (3+2\sqrt{2} \right)^n$$
as this would lead to $0$ appearing down the track of $U_n$ and the quotient is then invalid. This does not affect the final conclusion as for $U_0 = \frac{1}{2}$ we have 
$$ \frac{c}{d}=\left ( \frac{\sqrt{2}-1}{\sqrt{2}+1} \right )^n= \left (3-2\sqrt{2} \right)^n$$
