How to solve $P=\left(1+\frac{1}{3}\right)\left(1+\frac{1}{3^2}\right)\left(1+\frac{1}{3^3}\right)\ldots \infty$ How do I find the following product
$$P=\left(1+\frac{1}{3}\right)\left(1+\frac{1}{3^2}\right)\left(1+\frac{1}{3^3}\right)\ldots \infty$$

This is not an answer but I hope others will help to conclude something. 
First of all, take a logarithm from both sides:
$$
\log P=\log\prod_n\left(1+\frac{1}{3^n}\right)=\sum_n\log\left(1+\frac{1}{3^n}\right)
$$
Now using Taylor series, you get:
$$
\log(1+\frac{1}{3^m})=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} \frac{1}{3^{mn}}.
$$
Therefore we have:
$$
\log P=\log\prod_n\left(1+\frac{1}{3^n}\right)=\sum_{m=1}^\infty\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} \frac{1}{3^{mn}}\\
=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} \sum_{m=1}^\infty\frac{1}{3^{mn}}\\
=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} \frac{1}{3^{n}-1}.
$$
The preceding series is convergent. Now this is a Lambert series with $a_n=\frac{(-1)^{n+1}}{n}$ and $q=\frac{1}{3}$. For the moment, I am not sure if there is a closed form in terms of some special function or not.
Update 1: 
$$
\sum_{n=1}^\infty \frac{1}{3^{n}-1}=\frac{\psi_{\frac{1}{3}}(1)+\log\frac 23}{\log \frac 13}
$$
where $\psi_{q}(z)$ is q-Polygamma function.
