Evaluating $\prod_{n=1}^{\infty}\left(1+\frac{1}{2^n}\right)$ 
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? 
 A: 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 $$(1+\frac{1}{2^{2^k}})(1-\frac{1}{2^{2^k}})=1-\frac{1}{2^{2^{k+1}}}$$
A: $$\prod_{n\geq 1}\left(1+\frac{1}{2^n}\right)=\exp\sum_{n\geq 1}\sum_{m\geq 1}\frac{(-1)^{m+1}}{m 2^{mn}}=\exp\sum_{m\geq 1}\frac{(-1)^{m+1}}{m(2^m-1)} $$
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
$$\prod_{n\geq 1}\left(1+\frac{1}{2^n}\right)\approx 2.384231029.$$
A: 
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.
A: 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 $$\prod_{n\geq1}\left(1+\frac{1}{2^{n}}\right)=\left(-1;\frac{1}{2}\right)_{\infty}.$$
A: 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 $\approx 2.384$.
