jmc

We claim that $F(n)=2^n-\frac{1}{2^n}$ for all nonnegative integers $n.$ We prove this by strong induction.

The result for $n=0$ and $n=1.$ Assume that the result holds for $n=0,$ 1, 2, $\dots ,$ $k,$ for some nonnegative integer $k\ge 1,$ so $F(k-1)=2^{k-1}-\frac{1}{2^{k-1}}$ and $F(k)=2^k-\frac{1}{2^k}.$

Then $$\begin{array}{rl}F(k+1)& =\frac{5}{2}F(k)-F(k-1)\\ & =\frac{5}{2}\left(2^k-\frac{1}{2^k}\right)-\left(2^{k-1}-\frac{1}{2^{k-1}}\right)\\ & =\frac{5}{2}\cdot 2^k-\frac{5}{2}\cdot \frac{1}{2^k}-\frac{1}{2}\cdot 2^k+\frac{2}{2^k}\\ & =2\cdot 2^k-\frac{1}{2}\cdot \frac{1}{2^k}\\ & =2^{k+1}-\frac{1}{2^{k+1}}.\end{array}$$Thus, the result holds for $n=k+1,$ so by induction, the result holds for all $n\ge 0.$

Then the sum we seek is $$\sum _{n=0}^{\infty }\frac{1}{F(2^n)}=\sum _{n=0}^{\infty }\frac{1}{2^{2^n}-\frac{1}{2^{2^n}}}=\sum _{n=0}^{\infty }\frac{2^{2^n}}{(2^{2^n}{)}^2-1}.$$Let $x=2^{2^n}.$ Then $$\begin{array}{rl}\frac{2^{2^n}}{(2^{2^n}{)}^2-1}& =\frac{x}{x^2-1}\\ & =\frac{(x+1)-1}{x^2-1}\\ & =\frac{x+1}{x^2-1}-\frac{1}{x^2-1}\\ & =\frac{1}{x-1}-\frac{1}{x^2-1}\\ & =\frac{1}{2^{2^n}-1}-\frac{1}{2^{2^{n+1}}-1}.\end{array}$$Thus, our sum telescopes: $$\begin{array}{rl}\sum _{n=0}^{\infty }\frac{2^{2^n}}{(2^{2^n}{)}^2-1}& =\sum _{n=0}^{\infty }\left(\frac{1}{2^{2^n}-1}-\frac{1}{2^{2^{n+1}}-1}\right)\\ & =\left(\frac{1}{2^{2^0}-1}-\frac{1}{2^{2^1}-1}\right)+\left(\frac{1}{2^{2^1}-1}-\frac{1}{2^{2^2}-1}\right)+\left(\frac{1}{2^{2^2}-1}-\frac{1}{2^{2^3}-1}\right)+\cdots \\ & =\boxed{1}.\end{array}$$