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\[ \LARGE \prod_{n=0}^\infty \left (1 + \frac 1 {2^{2^n}} \right ) = \ ? \]

Note by Avn Bha
2 years, 5 months ago

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Multiply this product \(P\) by \(1\) in the form \(\dfrac{1 - \frac{1}{2}}{1 - \frac{1}{2}}.\) We then find that

\(P = 2*\left(1 - \dfrac{1}{2^{2^{1}}}\right)\left(1 + \dfrac{1}{2^{2^{1}}}\right)\left(1 + \dfrac{1}{2^{2^{2}}}\right)....\)

Now this product will continue to "compress", since \(\left(1 - \dfrac{1}{2^{2^{k}}}\right)\left(1 + \dfrac{1}{2^{2^{k}}}\right) = 1 - \dfrac{1}{2^{2^{k+1}}}\) for each successive \(k.\)

Thus in the limit as the upper index goes to infinity we see that \(P = 2.\)

Brian Charlesworth - 2 years, 5 months ago

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