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
$$P=\left(1+\frac{1}{3}\right)\left(1+\frac{1}{3^2}\right)\left(1+\frac{1}{3^3}\right)\ldots \infty$$
 A: For every $|x|\lt1$,
$$
\prod_{n=1}^\infty\left(1+x^n\right)=1+\sum_{n=1}^\infty a_nx^n,
$$
where
$$
a_n=\#\{I\subset\mathbb N\mid s(I)=n\},\qquad s(I)=\sum_{i\in I}i.
$$ 
Thus, $a_n$ is the number of partitions of $n$ into distinct parts, see page A000009 of the OEIS.
(Needless to say (or maybe one needs to say this after all), there exists no closed form formula for $a_n$ which would be valid for every $n$. Unless the value $x=\frac13$ would make the sum of the series simpler, there is no reason to expect a simple expression of the value of the infinite product.)
A: 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.
A: This would be too long for a comment I think. 
Anyways, lower and upper bounds for this product can be found very readily (how to find an exact value for the convergence, I don't know). 
The series presented is of the form (studied by Euler):
(1+x)(1+x^2)(1+x^3)(1+x^4)…
Now, if we were to expand this, imagine what the coefficient of x^1 term would be.
Well there is only 1 way you can choose it. Same goes for 1 (you multiply by 1 in every bracket). So the expansion would look something like this:
1 + x + x^2 + 2x^3 + 2x^4 +…
The 2 for x^3 comes from the fact that you can create x^3 in 2 ways: choose x in the first bracket, x^2 in the second - that is 1 way. Another way is to choose x^3 directly.
In general, the coefficients of this expansion are the partitions of n into distinct terms (as every power only appears once in the expansion). Also, each partition can only be selected in one way, as with each brackets the power of x^n always rises.  Take the distinct partitions of 6: 4 + 2, 5 + 1, 6 and 3 + 2 + 1. Things like 2 + 1 + 3 will not be counted by the coefficient ,as such a partition order cannot be selected from the expansion.The name for this property is: "order immaterial"
Now notice, the partition of a number n into distinct parts (order immaterial) for any given positive integer greater than n=1 is smaller than the general partition of n (order immaterial as well). (i.e. the "general" partitions (order immaterial) of 3, instead of just being 2 + 1, and 3, also include 1 + 1 + 1 So:
partition into distinct parts (order immaterial) < partition into parts (order immaterial) 
Now, clearly:
partition into parts (order immaterial) < partition into parts (order important)
as every immaterial partition, when order matters, can be created in multiple ways. I.e. instead of just 1 + 2 + 3, we have 1 + 3 + 2, 2 + 3 + 1, etc. Now for the partition into parts where order matters, the nth number has 2^(n-1) such partitions. This easily follows because:
n = 1 + 1 + 1 + 1 + 1 + …
Now every + can be converted into a "," This signifies that the 2 "1"s around the "," become a single number. So, for example:
n = 1 , 1 , 1 + 1 , 1  + 1 + 1
which represents 3 + 2 + 1 + 1.
Clearly here, all the possible sums (where order matters, will be selected). Now for a given n, there will be n-1 slots where we either choose "+" or ",". I.e. the number of choices are, 2^(n-1).
How does this relate to the original question.
Well, clearly by syllogism:
partition into distinct parts (order immaterial) <  partition into parts (order important)
So if we swap the coefficients of the original series with the coefficients for partition into parts (order important) and we see that the series still converges, it surely must also do so for the smaller distinct parts coefficients. 
Notice: 1 + 2*(1/3) + 2^2(1/3)^2 + 2^3(1/3)^3 + 2^4(1/3)^4 + … 
This is clearly a geometric series!
(2/3)^(n-1)
We know the sum of this as 1/(1-2/3), i.e. 3
Clearly, for this already enlarged version of the series, the sum to infinity converges. So, since the coefficients for a partition into distinct parts (order immaterial), in general are much smaller, the original series must also be converging to something!
Similarly, a lower bound can be established by considering no coefficients at all:
1 + 1/3 + 1/9 +…
which is equal to 1.5
Have a nice day :)
