Saudi Arabian Mathematical Competitions

We have for $k = 1, 2, \ldots$
$$2^{2^{k+1}} - 1 = \left(2^{2^{k}}\right)^{2} - 1 = \left(2^{2^{k}} - 1\right)\left(2^{2^{k}} + 1\right),$$
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
$$\begin{equation*} \frac{1}{f_{k}} = \frac{1}{x_{k}} - \frac{2}{x_{k+1}}, \quad k = 1, 2, \ldots \tag{1} \end{equation*}$$
where $x_{k} = 2^{2^{k}} - 1$. From (1) we get for $n = 1, 2, \ldots$
$$\begin{equation*} \frac{1}{f_{1}} + \frac{2}{f_{2}} + \frac{2^{2}}{f_{3}} + \ldots + \frac{2^{n-1}}{f_{n}} = \frac{1}{x_{1}} - \frac{2^{n}}{x_{n+1}} = \frac{1}{3} - \frac{2^{n}}{x_{n+1}} \tag{2} \end{equation*}$$
From (2) it follows
$$\frac{1}{f_{1}} + \frac{2}{f_{2}} + \frac{2^{2}}{f_{3}} + \ldots + \frac{2^{n-1}}{f_{n}} < \frac{1}{3}$$
for all positive integers $n$. Since
$$\lim_{n \rightarrow \infty} \frac{2^{n}}{x_{n+1}} = \lim_{n \rightarrow \infty} \frac{2^{n}}{2^{2^{n+1}} - 1} = \lim_{t \rightarrow \infty} \frac{t}{2^{2t} - 1} = 0$$
we obtain that the least real number $C$ with the desired property is $C = \frac{1}{3}$.