For $x 1,\lim_{n \rightarrow \infty}\mspace{2mu}\prod_{k = 0}^{n}\mspace{2mu}\left( 1 + \frac{2

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For $x > 1,\lim_{n \rightarrow \infty}\mspace{2mu}\prod_{k = 0}^{n}\mspace{2mu}\left( 1 + \frac{2}{x^{2^{k}} + x^{- 2^{k}}} \right)$ is equal to

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Sol. [Key Concept: $\prod(1 + \frac{1}{x^{2^{k}} + x^{{- 2}^{k}}})$ telescopes; take the limit.]$$\frac{\left( x^{1/2} + x^{- 1/2} ight)^{2}\left( x + x^{- 1} ight)^{2}\left( x^{2} + x^{- 2} ight)^{2}}{\left( x + x^{- 1} ight)\left( x^{2} + x^{- 2} ight)\left( x^{4} + x^{- 4} ight)}\ldots\frac{\left( x^{2^{n - 2}} + x^{- 2^{n - 2}} ight)^{2}}{x^{2^{n - 1}} + x^{- 2^{n - 1}}}\frac{\left( x^{2^{n - 1}} + x^{- 2^{n - 1}} ight)^{2}}{\left( x^{2^{n}} + x^{- 2^{n}} ight)}$$$$\begin{matrix} & \left( x^{1/2} + x^{- 1/2} ight)^{2}\left( x + x^{- 1} ight)\left( x^{2} + x^{- 2} ight)\left( x^{4} + x^{- 4} ight)\ldots\frac{\left( x^{2^{n - 1}} + x^{- 2^{n - 1}} ight)}{\left( x^{2^{n}} + x^{- 2^{n}} ight)} \ & \ = \frac{\left( x^{1/2} + x^{- 1/2} ight)^{2}\left( x + x^{- 1} ight)\left( x + x^{- 1} ight)\left( x^{2} + x^{- 2} ight)\left( x^{4} + x^{- 4} ight)\ldots\left( x^{2^{n - 1}} + x^{- 2^{n - 1}} ight)}{\left( x - x^{- 1} ight)\left( x^{2^{n}} + x^{- 2^{n}} ight)} \ & \ = \frac{\left( x^{1/2} + x^{- 1/2} ight)^{2}\left( x^{2^{n}} - x^{- 2^{n}} ight)}{\left( x - x^{- 1} ight)\left( x^{2^{n}} + x^{- 2^{n}} ight)} \ & \ \lim_{n ightarrow \infty}\mspace{2mu} p_{n} = \frac{\left( x^{1/2} + x^{- 1/2} ight)^{2}}{\left( x - x^{- 1} ight)} = \frac{(x + 1)^{2}}{x^{2} - 1} = \left( \frac{x + 1}{x - 1} ight) \end{matrix}$$

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