Saudi Arabia Mathematical Competitions

We have for $k=1,2,\dots$ $$2^{2^{k+1}}-1={\left(2^{2^k}\right)}^2-1=\left(2^{2^k}-1\right)\left(2^{2^k}+1\right),$$ hence we get $$\frac{1}{2^{2^{k+1}}-1}=\frac{1}{2}\left(\frac{1}{2^{2^k}-1}-\frac{1}{2^{2^k}+1}\right)$$ It follows $$\begin{array}{c}\frac{1}{f_k}=\frac{1}{x_k}-\frac{2}{x_{k+1}},k=1,2,\dots \end{array}\text{\hspace{1em}(1)}$$ where $x_k=2^{2^k}-1$. From (1) we get for $n=1,2,\dots$ $$\begin{array}{c}\frac{1}{f_1}+\frac{2}{f_2}+\frac{2^2}{f_3}+\dots +\frac{2^{n-1}}{f_n}=\frac{1}{x_1}-\frac{2^n}{x_{n+1}}=\frac{1}{3}-\frac{2^n}{x_{n+1}}\end{array}\text{\hspace{1em}(2)}$$ From (2) it follows $$\frac{1}{f_1}+\frac{2}{f_2}+\frac{2^2}{f_3}+\dots +\frac{2^{n-1}}{f_n}<\frac{1}{3}$$ for all positive integers $n$. Since $$\underset{n\to \infty }{\lim }\frac{2^n}{x_{n+1}}=\underset{n\to \infty }{\lim }\frac{2^n}{2^{2^{n+1}}-1}=\underset{t\to \infty }{\lim }\frac{t}{2^{2t}-1}=0$$ we obtain that the least real number $C$ with the desired property is $C=\frac{1}{3}$.