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I am trying to compute the infinite product

$$ \prod\limits_{n=2}^\infty \left(1+\frac{1}{2^n-2}\right) . $$

Wolfram Alpha says the result is $2$, but I can't seem to figure out why.

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HINT:

$$1+\frac1{2^n-2}=\frac{2^n-1}{2(2^{n-1}-1)}$$

Clearly, the denominator of each term gets cancelled by the numerator of the previous except for the last term

$$\implies\prod_{n=2}^N\left(1+\frac1{2^n-2}\right)=\frac{2^N-1}{2^{N-1}}$$

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Write

$$1 + \frac{1}{2^n - 2} = \frac{2^n - 1}{2^n - 2} = \frac{1}{2} \frac{2^n - 1}{2^{n-1} - 1}$$

so by telescoping,

$$\prod_{k = 2}^N \left(1 + \frac{1}{2^n - 2}\right) = \frac{1}{2^{N-1}}(2^N - 1) = 2 - \frac{1}{2^{N-1}}.$$

Since $\frac{1}{2^{N-1}} \to 0$ as $N \to \infty$, we have

$$\prod_{n = 2}^\infty \left(1 + \frac{1}{2^n-2}\right) = 2.$$

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