I would like to prove that the fractional part of $\sqrt{4^n-1}$ is greater than 1/2
for any natural n
.
Just to clarify- $frac(x)=$ $x$$-$$\lfloor x \rfloor$.
e.g frac($\pi$) = $\pi - 3$ .
My Attempts:
- Induction - for
n=1
$frac($$\sqrt{3}$$)>0.5$
Assuming $frac($$\sqrt{4^n-1}$$)>0.5,$ I could not prove that $frac($$\sqrt{4^m-1}$ $)>0.5.$$(m=n+1)$ - Declare the follow sequence $a$
n
= $frac($$\sqrt{4^n-1}$
And prove it's limit is 1 (I believe it converges to 1), I tried to prove it using the limit's definition but again I got stuck.
Would love to get some help with my attempts here or to hear of a different method. Thanks in advance.