I attempted to prove the following but am unsure if my logic is solid.
Define the sequence of numbers $A_i$, by
$A_0 = 2$
$A_{n+1} = \frac{A_n}{2} + \frac{1}{A_n}$ for $n\geq 1$
Prove that $A_n \leq\sqrt2 + \frac{1}{2^n}$ for all $n\geq0$.
proof attempt: (using strong induction)
assume $A_n$ is true for $A_0$, $A_1$, ... $A_n$,
prove $A_{n+1} \leq\sqrt2 + \frac{1}{2^{n+1}}$
substitute $A_n$ into recursive definition:
$A_{n+1} = \frac{\sqrt2 + \frac{1}{2^n}}{2} + \frac{1}{\sqrt2 + \frac{1}{2^n}}$
$\frac{\sqrt2 + \frac{1}{2^n}}{2} + \frac{1}{\sqrt2 + \frac{1}{2^n}} \leq \sqrt2 + \frac{1}{2^{n+1}}$
This is where I am unsure if my thinking is correct. I attempted to take the limit as n -> infinity such that all the 1/2^n terms go to 0.
We are left with...
$\frac{\sqrt{2}}{2} +\frac{1}{\sqrt2} \leq\sqrt2$
which gives
$\frac{2}{\sqrt2}\leq\sqrt2$,
which is true...but I am not sure if I just proved it for all n or just very large n.
Could somebody offer some feedback? Thank you.