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

Question

Compute:$$\prod_{n=1}^{\infty}\left(1+\frac{1}{2^n}\right)$$

I and my friend came across this product. Is the product till infinity equal to $1$?

If no, what is the answer?

Answer 1

Is the product till infinity equal to $1$?

Certainly not! All the individual terms are greater than $1$. So if you multiply them together, you will always be increasing and cannot get back to $1$.

If no, what is the answer?

The product in question is $$ \prod_{n=1}^\infty (1 + x^n) $$ where $x = \frac12$. This product equals $$ \sum_{n=0}^\infty q(n) x^n $$ where $q(n)$ is the number of partitions of $n$ into distinct parts (each part $\ge 1$), and also equals $$ \prod_{n=1}^\infty \frac{1}{1 - x^{2n-1}} = \frac{\Phi(x^2)}{\Phi(x)} $$ (see Wikipedia), where here $\Phi$ is the Euler function, not to be confused with Euler's totient function. So your product is equal to $$ \boxed{\frac{\Phi(1/4)}{\Phi(1/2)} = 2.38423\ldots}. $$ I don't expect this can be simplified.

Answer 2

In the link you provided, the product seems to be $$\prod_{n=0}^{\infty} \left (1+\frac{1}{2^{2^n}}\right)$$

Note that $$(1-\frac{1}{2})\prod_{n=0}^{\infty} (1+\frac{1}{2^{2^n}})=\lim_{n \to \infty} 1-\frac{1}{2^{2^{n+1}}}=1$$ From the fact that $$(1+\frac{1}{2^{2^k}})(1-\frac{1}{2^{2^k}})=1-\frac{1}{2^{2^{k+1}}}$$

Answer 3

$$\prod_{n\geq 1}\left(1+\frac{1}{2^n}\right)=\exp\sum_{n\geq 1}\sum_{m\geq 1}\frac{(-1)^{m+1}}{m 2^{mn}}=\exp\sum_{m\geq 1}\frac{(-1)^{m+1}}{m(2^m-1)} $$ where the last series is (rapidly) convergent by Leibniz' test. It follows that the original product is between $\exp\left(\frac{121}{140}\right)$ and $\exp\left(\frac{3779}{4340}\right)$. By truncating the shown series at $m=30$ we get $$\prod_{n\geq 1}\left(1+\frac{1}{2^n}\right)\approx 2.384231029.$$

Answer 4

Another way to obtain the same answer of 6005. If we take the log of this product and the Taylor expansion of log we get $$\sum_{n\geq1}\sum_{k\geq1}\frac{\left(-1\right)^{k+1}}{k2^{mk}}=\sum_{k\geq1}\frac{\left(-1\right)^{k}}{k\left(1-2^{k}\right)}=-\sum_{k\geq1}\frac{\left(-1\right)^{k}}{k2^{k}\left(1-\frac{1}{2^{k}}\right)} $$ $$=-\sum_{k\geq1}\frac{1}{k4^{k}\left(1-\frac{1}{4^{k}}\right)}+\sum_{k\geq1}\frac{1}{k2^{k}\left(1-\frac{1}{2^{k}}\right)} $$ and since we have $$\log\left(\Phi\left(q\right)\right)=-\sum_{k\geq1}\frac{q^{n}}{k\left(1-q^{n}\right)} $$ where $\Phi\left(q\right) $ is the Euler's function we have $$\sum_{n\geq1}\sum_{k\geq1}\frac{\left(-1\right)^{k+1}}{k2^{mk}}=\log\left(\frac{\Phi\left(1/4\right)}{\Phi\left(1/2\right)}\right). $$ As wrote by 6005, probably there is no simplification for this result. The result can be written also as a q-Pochhammer symbol $$\prod_{n\geq1}\left(1+\frac{1}{2^{n}}\right)=\left(-1;\frac{1}{2}\right)_{\infty}.$$

Answer 5

This is not a full answer. There are already full answers posted.

I can prove that the sequence $x_n=\prod_{i=1}^n\left(1+\frac{1}{2^i}\right)$ converges, is strictly increasing and less than $e$, so $\prod_{i=1}^{+\infty}\left(1+\frac{1}{2^i}\right)\le e$. Notice that $x_n>0$, $\forall n$, so $x_{n+1}=x_n\left(1+\frac{1}{2^n}\right)>x_n$, $\forall n$. $$x_1=1.5\le x_n=e^{\ln\prod_{i=1}^{n}\left(1+\frac{1}{2^i}\right)}=$$

$\ln\left(1+\frac{1}{n}\right)<\frac{1}{n}$, $\forall n\ge 1$, $n\in\mathbb Z$ is true. A more general inequality $e^x> x+1$, $\forall x\in\mathbb R$, $x\neq 0$ is also true.

$$=e^{\sum_{i=1}^n \ln\left(1+\frac{1}{2^i}\right)}<e^{\sum_{i=1}^n \frac{1}{2^i}}<e^{\sum_{i=1}^{+\infty}\frac{1}{2^i}}=e$$

WolframAlpha says that the infinite product is $\approx 2.384$.