How to prove that frac($\sqrt{4^n-1}$) is greater than 1/2 for any natural n

Question

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:

1. 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)$
   
2. 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.

Answer 1

Note that $\sqrt {4^n-1}$ is very close to $\sqrt {4^n} = 2^n$. If we can somehow show that $\sqrt {4^n-1} > 2^n-\frac12$ then $\operatorname {frac} (\sqrt{4^n-1}) > \frac12$. But

$$\left(2^n-\frac12\right)^2 = 4^n-2^n+\frac14 < 4^n-2^n+1 \le 4^n-1$$

The last inequality holds true for $n \ge 1$.

Answer 2

Note that$$2^n-\sqrt{4^n-1}=\frac1{2^n+\sqrt{4^n-1}}<\frac1{2^n}\leqslant\frac12.$$Can you take it from here?