Evaluating a product $\prod_{n=0}^{100}\left(1+\dfrac{1}{a^{2^n}}\right)$

Question

How would you evaluate this product? $$\displaystyle \prod_{n=0}^{100}\left(1+\dfrac{1}{a^{2^n}}\right)$$ I know one way in which multiplying the product by $\left(1-\dfrac{1}{a}\right)$ does the job but I was looking for a more formal approach.

Answer 1

If you want to keep the product sign: $$\begin{align}\left(1-\frac1a\right)\prod_{n=0}^{100}\left(1+\frac1{a^{2^n}}\right)&=\left(1-\frac1{a^2}\right)\prod_{n=1}^{100}\left(1+\frac1{a^{2^n}}\right)\\&=\left(1-\frac1{a^{2^2}}\right)\prod_{n=2}^{100}\left(1+\frac1{a^{2^n}}\right)\\&=\cdots\\&=\left(1-\frac1{a^{2^{100}}}\right)\prod_{n=100}^{100}\left(1+\frac1{a^{2^n}}\right)\\&=\left(1-\frac1{a^{2^{100}}}\right)\left(1+\frac1{a^{2^{100}}}\right)=1-\frac1{a^{2^{101}}}\end{align}$$

Answer 2

$$1 + {a^{-2^n}} = \frac{1 - a^{-2^{n+1}}}{1 - a^{-2^n}}$$ So we have a telescoping product $$ \prod_{n=0}^N \left(1 + a^{-2^n}\right) = \frac{1-a^{-2^{N+1}}}{1-a^{-1}}$$

Answer 3

Isn't there a nice way to do this keeping the Pi sign?

Let $\,b = \dfrac{1}{a}\,$, then using the algebraic identity $\,1+b^{2^k}=\dfrac{1-b^{2^{k+1}}}{1-b^{2^k}}\,$:

$$\require{cancel} \prod_{n=0}^{100}\left(1+b^{2^n}\right) = \prod_{n=0}^{100} \frac{1-b^{2^{n+1}}}{1-b^{2^n}} = \frac{\prod_{\color{blue}{n=0}}^{\color{blue}{100}} \left(1-b^{2^{\color{blue}{n+1}}}\right)}{\prod_{n=0}^{100} \left(1-b^{2^{n}}\right)}=\frac{\prod_{\color{red}{n=1}}^{\color{red}{101}} \left(1-b^{2^{\color{red}{n}}}\right)}{\prod_{n=0}^{100} \left(1-b^{2^{n}}\right)} \\[10px] =\frac{\left(1-b^{2^{101}}\right) \cdot \bcancel{\prod_{n=1}^{100} \left(1-b^{2^{n}}\right)}}{\left(1-b^{2^{0}}\right) \cdot \bcancel{\prod_{n=1}^{100} \left(1-b^{2^{n}}\right)}} = \ldots $$

Answer 4

An interesting way to look at this product is by realizing that it is tied to the binary representation of the integers spanning from $0$ to $2^{N+1}-1$.

In fact, since $(1+x)(1+y)=1+x+y+xy$, then it is easy to deduct that $${\prod _{n=0}^N \left(1+x^{2^n}\right)}={\sum_{0 \le k \le 2^{N+1}-1}x^k}$$ i.e., where $k$ spans all the integers representable by $N$ bits.