If $U_0 = 0$ and $U_n=\sqrt{U_{n-1}+(1/2)^{n-1}}$, then $U_n < U_{n-1}+(1/2)^n$ for $n > 2$ Letting $$U_n=\sqrt{U_{n-1}+(1/2)^{n-1}}$$ where $U_0=0$, prove that:
$$U_n < U_{n-1}+(1/2)^n$$ where $n>2$
 A: Here is a useful factoid:

For every $x\geqslant1$ and $y\gt0$, $\sqrt{x+2y}\lt x+y$.

Now, apply this to your setting. First note that $U_n\geqslant1$ implies $U_{n+1}\geqslant1$. Since $U_1=1$, this proves that $U_n\geqslant1$ for every $n\geqslant1$. Then, choosing $n\geqslant2$, $x=U_{n-1}$ and $y=1/2^n$, the factoid yields $U_n\lt U_{n-1}+1/2^n$, as desired.
Finally the result holds for every $n\geqslant2$ (and not only $n\gt2$).
Can you prove the factoid above?
A: Observe that $ U_1 = 1,\ U_2 = \sqrt{1+(1/2)}> 1 $ and $ U_n > 1 $ implies $U_{n+1} > 1$ using the recursion, hence inductively $ U_n > 1 $ for all $ n>1 $.
Now rationalizing suitably you will find 
$$ U_k - U_{k-1} = \sqrt{U_{k-1} + (1/2)^{k-1}}-\sqrt{U_{k-2}+(1/2)^{k-2}} \\
 = \frac{U_{k-1}-U_{k-2} -(1/2)^{k-1}}{\sqrt{U_{k-1} + (1/2)^{k-1}}+\sqrt{U_{k-2}+(1/2)^{k-2}}} < \frac{U_{k-1}-U_{k-2}}{U_k + U_{k-1}} < \frac{U_{k-1}-U_{k-2}}{2} $$
So you have $ U_1-U_0 = 1 $ and for all $ k > 1\ $
$ \frac{U_k-U_{k-1}}{U_{k-1}-U_{k-2}} < (1/2) $ So multiplying terms for $ 2 \leq k \leq n $ you have for $ n \geq 2 $
$$ U_n - U_{n-1} < (1/2)^n $$
