Let $f_{n}(x) = \frac{x}{1 - x^{2}} + \frac{x^{2}}{1 - x^{4}} + \frac{x^{4}}{1 - x^{8}} + \ldots + \

Question

Let $f_{n}(x) = \frac{x}{1 - x^{2}} + \frac{x^{2}}{1 - x^{4}} + \frac{x^{4}}{1 - x^{8}} + \ldots + \frac{x^{2^{n - 1}}}{1 - x^{2^{n}}}$ where $x > 0,x 
eq 1$ and $f(x) = \lim_{n ightarrow \infty}\mspace{2mu} f_{n}(x)$, then

Accepted Answer

Sol. [Key Concept: Use the definition to test each option.]$$f_{n}(x) = \frac{1}{2}\left\lbrack \frac{1}{1 - x} - \frac{1}{1 + x} + \frac{1}{1 - x^{2}} - \frac{1}{1 + x^{2}} + \frac{1}{1 - x^{4}} - \frac{1}{1 + x^{4}} + \ldots ightbrack$$$$\begin{matrix} = & \frac{1}{2}\left\lbrack - \frac{1}{1 - x} - \frac{1}{1 + x} - \frac{1}{1 + x^{2}} - \frac{1}{1 + x^{4}} - \ldots. ight.\ \ & \left. \ - \frac{1}{1 + x^{2n - 1}} + \frac{2}{1 - x} + \frac{1}{1 - x^{2}} + \frac{1}{1 - x^{4}} + \ldots + \frac{1}{1 - x^{2^{n - 1}}} ightbrack \ = & \frac{1}{2}\left\lbrack - \frac{1}{1 - x^{2}} - \frac{1}{1 + x^{2}} - \frac{1}{1 + x^{4}} - \ldots. ight.\ \ 	herefore\ & \frac{1}{2}\left\lbrack \frac{1}{1 - x} - \frac{1}{1 - x^{2^{n - 1}}} + \frac{2}{1 - x} + \frac{1}{1 - x^{4}} + \frac{1}{1 - x^{8}} + \ldots + \frac{1}{1 - x^{2^{n - 1}}} ightbrack = \frac{1}{1 - x} - \frac{1}{1 - x^{2^{n}}} \ 	herefore\ & \ (x) = \left\{ \begin{matrix} \frac{x}{1 - x} & x \in (0,1) \ \frac{1}{1 - x} & x \in (1,\infty) \end{matrix} ight.\ \end{matrix}$$✔ (a) $f(\frac{14) =}{\frac{1/4}{3/4}} = \frac{13}{}$✔ (b) $f(\frac{12) = 1}{}$✔ (c) $f(3) = \frac{1}{1 - 3} = - \frac{12}{}$✘ (d) $f(5) = - \frac{14}{
eq} - \frac{16}{}$

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