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How can I calculate the following product of series? $$ \prod_{k=0}^{n-1}\left (1+\frac{1}{2^{2^k}}\right) $$....Can I take a geometric series and compare it with that?

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Multiply through by $\frac12=1-\frac1{2^{2^0}}$ to get

$$\begin{align}\left(1-\frac1{2^{2^0}}\right)\prod_{k=0}^n\left(1+\frac1{2^{2^k}}\right)&=\left(1-\frac1{2^{2^0}}\right)\left(1+\frac1{2^{2^0}}\right)\prod_{k=1}^n\left(1+\frac1{2^{2^k}}\right)\\&=\left(1-\frac1{2^{2^1}}\right)\left(1+\frac1{2^{2^1}}\right)\prod_{k=2}^n\left(1+\frac1{2^{2^k}}\right)\\&=\left(1-\frac1{2^{2^2}}\right)\left(1+\frac1{2^{2^2}}\right)\prod_{k=3}^n\left(1+\frac1{2^{2^k}}\right)\\&=\left(1-\frac1{2^{2^3}}\right)\left(1+\frac1{2^{2^3}}\right)\prod_{k=4}^n\left(1+\frac1{2^{2^k}}\right)\\&=\quad\vdots\\&=\left(1-\frac1{2^{2^{n+1}}}\right)\end{align}$$

$$\prod_{k=0}^n\left(1+\frac1{2^{2^k}}\right)=2-\frac1{2^{2^{n+1}-1}}$$

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As a hint, consider $n = 3$:

$\displaystyle\prod_{k = 0}^{2}\left(1+\dfrac{1}{2^{2^k}}\right) = \left(1+\dfrac{1}{2^1}\right)\left(1+\dfrac{1}{2^2}\right)\left(1+\dfrac{1}{2^4}\right)$

$= 1 + \dfrac{1}{2^1} + \dfrac{1}{2^2} + \dfrac{1}{2^1} \cdot \dfrac{1}{2^2} + \dfrac{1}{2^4} + \dfrac{1}{2^1} \cdot \dfrac{1}{2^4} + \dfrac{1}{2^2} \cdot \dfrac{1}{2^4} + \dfrac{1}{2^1} \cdot \dfrac{1}{2^2} \cdot \dfrac{1}{2^4}$

$= 1 + \dfrac{1}{2^1} + \dfrac{1}{2^2} + \dfrac{1}{2^3} + \dfrac{1}{2^4} + \dfrac{1}{2^5} + \dfrac{1}{2^6} + \dfrac{1}{2^7}$

$= 2-\dfrac{1}{2^7}$

Can you generalize this for larger values of $n$?

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  • $\begingroup$ I'd love to see the full proof, as multiplying products out is quite a tedious thing. $\endgroup$ – Simply Beautiful Art May 31 '17 at 19:53
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This is really the same as JimmyK4542's answer, but written in the general case.

If you expand the product, you find that it is the sum of all possible terms of the form $$\frac{1}{2^{a_0 + a_1 2 + a_2 2^2 + \dots + a_{n-1}2^{n-1}}}$$ where each $a_i$ is either $0$ or $1$. Viewing the exponents as integers written in binary, they run though all integers $j = 0, 1, 2, \dots 2^n - 1$. We therefore find that the product is equal to $$\sum_{j = 0}^{2^n - 1} \frac{1}{2^j} = 2 - \frac{1}{2^{2^n - 1}}.$$

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    $\begingroup$ why is this not getting many upvotes, that was genius $\endgroup$ – Oussama Boussif Jun 1 '17 at 2:20
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Partial Answer:

If I were you, I would first simplify that to $$1+\frac{1}{2^{2^k}}=\frac{2^{2^k}+1}{2^{2^k}}$$ and so we have $$\prod_{k=1}^{n-1}\frac{2^{2^k}+1}{2^{2^k}}$$ By splitting this up, we get $$\prod_{k=1}^{n-1}\frac{1}{2^{2^k}}*\prod_{k=1}^{n-1}\big(2^{2^k}+1\big)$$ Your insight about there being geometric series is correct, because look what happens when we evaluate the first product: $$\frac{1}{2^{2^1}}*\frac{1}{2^{2^2}}*...*\frac{1}{2^{2^{n-1}}}$$ $$\frac{1}{2^{2^1+2^2+...+2^{n-1}}}$$ $$\frac{1}{2^{2^n-1}}$$ So we have $$\frac{1}{2^{2^n-1}}*\prod_{k=1}^{n-1}\big(2^{2^k}+1\big)$$ I'm afraid that's all I've got... I have no idea how to evaluate that last part.

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  • $\begingroup$ 1) the sum of the geometric series would be $2^{2^n-1}$ if the product went from $0$ (not 1) to $n-1$; 2) to evaluate the last part, it should be followed Simply Beautiful Art's method: to multiply by $2^{2^0}-1$ and it will boil down to $2^{2^n}-1.$ $\endgroup$ – farruhota Jun 1 '17 at 10:52

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