# how do we solve this ?

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

Note by Avn Bha
3 years, 1 month 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 $$P = 2.$$

- 3 years, 1 month ago