Let $f(x) = \lim_{n \rightarrow \infty}\mspace{2mu}\left( \frac{1}{x + 1} + \frac{2}{x^{2} + 1} + \f

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Let $f(x) = \lim_{n \rightarrow \infty}\mspace{2mu}\left( \frac{1}{x + 1} + \frac{2}{x^{2} + 1} + \frac{4}{x^{4} + 1} + \ldots.. + \frac{2^{n}}{x^{2^{x}} + 1} \right)$ for $x > 1$, then

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Sol. [Key Concept: Establish the sum/product identities for the defined sequence.]$$\frac{1}{x - 1} + \left( \frac{1}{x + 1} - \frac{1}{x - 1} ight) + \frac{2}{x^{2} + 1} + \frac{4}{x^{4} + 1} + \ldots. + \frac{2^{n}}{x^{2^{n}} + 1}$$$$\begin{matrix} & \ = \frac{1}{x - 1} + 2\left( \frac{1}{x^{2} + 1} - \frac{1}{x^{2} - 1} ight) + \frac{4}{x^{4} + 1} + \ldots.. + \frac{2^{n}}{x^{2^{n}} + 1} \ & \ = \frac{1}{x - 1} + 4\left( \frac{1}{x^{4} + 1} - \frac{1}{x^{4} - 1} ight) + \frac{8}{x^{8} + 1} + \ldots.. + \frac{2^{n}}{x^{2^{n}} + 1} = \frac{1}{x - 1} - \frac{2^{n + 1}}{x^{2^{n + 1}} - 1} \ \Rightarrow \ f(x) & \ = \frac{1}{x - 1}\forall x > 1 \end{matrix}$$✘ (a) with $f(x) = \frac{1}{x - 1}$, $\sum_{r = 2}^{6}(r - 1) = 15 
eq 20$✘ (b) $f(5) = \frac{1}{4} 
eq \frac{16}{}$✔ (c) $f(5) = \frac{1}{5 - 1} = \frac{14}{}$✔ (d) $\sum_{r = 2}^{6}\frac{1}{f(r)} = \sum_{r = 2}^{6}(r - 1) = 15$

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