Does $\prod\limits_{n=1}^{\infty} \left(1+\frac{1}{3^{2n-1}}\right)$ have a closed result? Does $\prod\limits_{n=1}^{\infty} \left(1+\dfrac{1}{3^{2n-1}}\right)$ has a closed result?
I have computed it by WolframAlpha and got a numerical solution $1.564934\cdots$.
 A: This is not the result Wolfram Alpha gave me (have a look here.
The result, as written in the page is
$$\frac{1}{4}\left(-3;\frac{1}{9}\right)_{\infty }$$ where appears the Pochhammer symbol.
Its decimal representation is not recognized by inverse symbolic calculators.
A: One can derive the following result using complex analysis:
$$\prod_{n=1}^{\infty} \left(1+\dfrac{1}{3^{2n-1}}\right) = 3^{-\frac{1}{24}}\exp\left(\frac{\pi^2}{24\log(3)}\right)\prod_{n=1}^{\infty}\left(1+\dfrac{1}{a^{2n-1}}\right)\tag{1}$$
where
$$a = \exp\left(\frac{\pi^2}{\log(3)}\right)\approx 7972.08287589$$
So, it's not a closed form result, but note that because $a\gg 3$, the product on the r.h.s. converges much faster than the original product.
To derive this result, take the logarithm of the product to turn it into a summation:
$$S = \sum_{n=1}^{\infty} \log\left(1+\dfrac{1}{3^{2n-1}}\right) $$
Expand the logarithm in series, exchange the two summations and perform the inner summation:
$$S = \sum_{k=1}^{\infty}\frac{(-1)^{k+1}3^k}{k}\sum_{n=1}^{\infty}3^{-2nk}= \sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k\left(3^k-3^{-k}\right)}$$
We can write this a a summation over all integers except zero, as the summand is an even function:
$$S = \frac{1}{4}\sum_{k\neq 1}^{\infty}\frac{(-1)^{k+1}}{k\sinh(\log(3) k)}$$
Alternating summations over integers of a meromorphic functions $f(z)$ can sometimes be computed by considering the contour integral of $\dfrac{\pi f(z)}{\sin(\pi z)}$ over a contour of radius $R$ around the origin. If in the limit of $R\to \infty$ the contour integral vanishes, then the sum of all residues must be zero. The summation is (in our case) up to a minus sign given by the sum of the residues at the poles of $\dfrac{1}{\sin(\pi z)}$.
It's easily verified that the contour integral will indeed tend to zero for $R\to \infty$, therefore $S$ is given by the sum of the residues at zero and at the zeroes of $\sinh(\log(3) z)$ of the function:
$$g(z) = \frac{\pi}{4 z\sin(\pi z)\sinh(\log(3) z)}$$ 
The residue at zero is:
$$\frac{\pi^2 - \log^2(3)}{24\log(3)}$$
The sum of the residues at the zeroes of  $\sinh(\log(3) z)$  is:
$$\frac{1}{4}\sum_{k\neq 1}^{\infty}\frac{(-1)^{k+1}}{k\sinh\left(\dfrac{\pi^2 k}{\log(3)}\right)}$$
Exponentiating the sum of the residues yields (1).
