computing the product $\prod_{n=1}^{2016} \frac{2^{2^{n-1}}+1}{2^{2^{n-1}}}$

Question

how can i calculate the product: $\prod_{n=1}^{2016} \frac{2^{2^{n-1}}+1}{2^{2^{n-1}}}$?

I can see that in the denominator it's a geometric series, but in the numerator i can't see how to simplify.

Thanks.

Answer 1

Interesting problem!

You can write this as : $\prod_{n=1}^{2016} \left ( 1 + \frac{1}{2^{2^{n-1}}} \right )$

If we write $\frac12$ as $a$, then the problem is $$\prod_{n=1}^{2016} \left ( 1 + a^{2^{n-1}} \right )$$ If you observe carefully, every term of the form $a^k$ occurs exactly once in the above product for all $0 \leq k \leq {2^{2016}-1}$. Think of the product as $(a^0 + a^{2^0})(a^0 +a^{2^1})(a^0 + a^{2^2})(a^0+a^{2^3})...$. Then for any number $1 \leq k < 2^{2016}$, consider its binary representation. If the $i^{th}$ bit is $1$, then, for $a^k$ choose $a^{2^i}$ from the corresponding bracket, else choose $a^0$ from that bracket. Hence, every number of the form $a^k$ can be seen to occurs exactly once in the above expression. As for $a^0$, just choose $a^0$ from all the brackets.

Thus, it is effectively a geometric progression. $$\therefore \prod_{n=1}^{2016} \left ( 1 + \frac{1}{2^{2^{n-1}}} \right ) = \frac{1-\frac1{2^{2^{2016}}}}{1-\frac12} = 2 - \frac{2}{2^{2^{2016}}}$$

Answer 2

Note that \begin{align*} \prod_{n=1}^{2016} \frac{2^{2^{n-1}}+1}{2^{2^{n-1}}}& =\frac{(2^{2^{0}}-1)(2^{2^{0}}+1)\cdots (2^{2^{2015}}+1)}{(2^{2^{0}}-1)2^{2^{0}}\cdots 2^{2^{2015}}}\\ & =\frac{(2^{2^{1}}-1)(2^{2^{1}}+1)\cdots (2^{2^{2015}}+1)}{(2^{2^{0}}-1)2^{2^{0}+\cdots +2^{2015}}}\\ & =\frac{(2^{2^{2}}-1)(2^{2^{2}}+1)\cdots (2^{2^{2015}}+1)}{2^{2^{0}+\cdots +2^{2015}}}\\ & \cdots\\ & =\frac{2^{2^{2016}}-1}{2^{2^{2016}-1}} \end{align*}

Answer 3

Let $a_n=2^{(2^n)}+1$ be the $n$-th Fermat number. Since: $$ a_n-2 = 2^{2^n}-1 = \left(2^{2^{n-1}}-1\right)\cdot \left(2^{2^{n-1}}+1\right)\tag{1} $$ we get, by induction, $$ \prod_{n=0}^{N} a_n = a_{N+1}-2 \tag{2} $$ hence: $$ \prod_{n=1}^{2016}\frac{2^{2^{n-1}}+1}{2^{2^{n-1}}} = \frac{2^{2^{2016}}-1}{2^{2^{2016}-1}}=\color{red}{2-\frac{2}{2^{2^{2016}}}}.\tag{3}$$