An easy invariant,just note that the sum of all reciproal values increases(this is true and after some short calculation it reduses to prove that (a-b)^2>=0),so we only need to prove that 1/1 +1/2 + 1/3 + ...+1/4^n > n,but this is easy since 1 + 1/2 + 1/3 + 1/4 > 1, 1/5 + 1/6 + 1/7 + 1/8 > 1/2,...1/2^k+1 + ...+1/2^(k+1)>2^k*1/2^(k+1)=1/2,so the conclusion follows,so we are finished.

Define $s$ as the final number. by $Q.M. \geq A.M.:$ $\sqrt{\dfrac{\dfrac{1}{a^2}+\dfrac{1}{b^2}}{2}}\geq\dfrac{\dfrac{1}{a}+\dfrac{1}{b}}{2} \implies \sqrt{\dfrac{2a^2+2b^2}{a^2b^2}} \geq \dfrac{1}{a}+\dfrac{1}{b} \implies \bigg(\dfrac{ab}{\sqrt{2a^2+2b^2}}\bigg)^{-1} \geq a^{-1}+b^{-1}$. By this, we can say that the sum of the inverses always increase, so: $\dfrac{1}{s} \geq 1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{4^n}>\ln{4^n+1}>\ln{4^n}=n \cdot \ln{4}> n$ So, $\dfrac{1}{s}>n \implies \dfrac{1}{n}>s$.