How to establish inequality $U_{n+1} \leq U_{n}+\frac{1}{2^{^n}}$ $\begin{cases}
U_{1} = 1
      \\
U_{n+1} = \sqrt{U_{n}²+\frac{1}{2^{^n}}}
   \end{cases}$
How do I establish the following inequality ?
For $n \geq 1$
$U_{n+1} \leq U_{n}+\frac{1}{2^{^n}}$
I thought about proving that the sequence is increasing so that
$U_{n+1} \geq U_{n}$
But it's pretty useless here, I don't know how to advance any further.
 A: $ U_{n+1} \leq U_n + \frac{1}{2^n} \iff \sqrt{U_n^2 + \frac{1}{2^n}} \leq U_n + \frac{1}{2^n}. $
Squaring both sides, since $U_n$ is clearly positive we want to show that
$$U_n^2 + \frac{1}{2^n} \leq U_n^2 + \frac{U_n}{2^{n-1}} + \frac{1}{4^n}$$
Now it's enough to show that $U_n \geq 1$ for all $n$,  because then we have $\frac{U_n}{2^{n-1}} \geq \frac{1}{2^{n-1}} \geq \frac{1}{2^n}$ so we're done.
$U_n \geq 1$ should be obvious. (Turns out that $U_n$ increasing isn't useless!)
A: $\begin{cases}
u_{1} = 1
      \\
u_{n+1} = \sqrt{u_{n}^2+\frac{1}{2^{^n}}}
\end{cases}
$
Therefore
$u_{n+1}^2
=u_n^2+\dfrac1{2^n}
$
so
$u_{n+1}^2-u_n^2
=\dfrac1{2^n}
$
so
$u_n > 1$
for $n > 1$.
Therefore
$u_{n+1}-u_n
=\dfrac1{2^n(u_{n+1}+u_n)}
\lt\dfrac1{2^n2}
=\dfrac1{2^{n+1}}
$.
A: A comment:
One can prove that
$$U_{n+1} \leq U_{n}+\frac{1}{2^{\frac{n}{2}}} \qquad (1)$$
Simply
$$U^2_{n+1} \leq U^2_{n}+\frac{1}{2^{^n}}\leq U^2_{n}+\frac{1}{2^{\frac{n}{2}-1}}U_{n}+\frac{1}{2^{^n}}=(U_{n}+\frac{1}{2^{\frac{n}{2}}})^{2}$$
Is (1) enough for your purposes ?
