How to find the product $\Pi_{n=1 }^\infty (1+x^{2n-1 })$? $\Pi_{n=1 }^\infty (1+x^{2n-1 })$ 
Definition: $\Pi_{k=1}^\infty f(k)$ is said to converge to $l\in \mathbb R-\{0\}$ if the limit $\Pi_{k=1}^n f(k)\to l$ as $n\to \infty$ and we write $\Pi_{n=1 }^\infty (1+x^{2n-1 })=l$. 
I tried to find the product as below but got stuck.
For brevity let's say $a_n=1+x^{2n-1 }$. Now for $\Pi_{k=1}^\infty a_k$ to converge we must have $a_n=\frac{\Pi a_n}{\Pi a_{n-1}}\to 1$. It follows that $|x|\lt 1$. 
Let $P_n=\Pi_{k=1}^n a_k$. It follows that $\frac {\ln P_n}{n}\to 0$ ($n$th term of a sequence tends to $0\implies $ arithmetic means of $n$ elements converges to $0$ ). 
$\therefore$ Given any $\epsilon\gt 0$, we'll have $|\ln P_n|\lt \epsilon n$ for sufficiently large values of $n$ and then $e^{-\epsilon n}\lt P_n\lt e^{\epsilon n}$. Stuck.
How do I proceed from here? Thanks.
 A: Here is how you show convergence:

*

*$\prod_{k=1}^\infty(1+x^{2k-1})$ converges if and only if $\sum_{k=1}^\infty\mathrm{log}(1+x^{2k-1})$ converges.


*for all positive $x$, you have the bound $|\log(1+x^{2k-1})| \le |x^{2k-1}|$


*$\sum_k x^{2k-1}$ is (absolutely) convergent for all $|x|<1$ (just a geometric series). So the original series converges too.
For $|x|\ge1$ you already showed that the product cannot possibly converge. Which leaves open the case $x\in(-1,0)$. I think the most elegant thing to do here so to rewrite the original product into a power-series. Than you know that it must have a convergence radius equal in the positive and negative direction. It follows that the original expression converges exactly if $|x|<1$.
For the value it converges to, look for 'Pochhammer Symbol' (as @JJacquelin noted).
A: $$
f(x) := \prod_{n=1 }^\infty (1+x^{2n-1 }),\qquad x \in \mathbb C, |x| < 1
$$
Converges to a holomorphic function.  This is not an "elementary function".  It can be written in terms of some known special functions.
added
For example, the Dedekind eta function is
$$
\eta(q) = q^{1/24}\prod_{n=1}^\infty(1-q^n)
$$
This means
$$
f(x) = \frac{\eta(-x)}{\eta(x^2)}\;(-x)^{1/24}
$$
Some values of $\eta$ are known in closed form.  So we can compute some values of $f(x)$.  Examples (writing $q=-x, q = e^{2\pi i \tau}$):
With $\tau = \frac{i}{2}$ we get
$$
f(-e^{-\pi}) = 2^{1/8} e^{-\pi/24} \approx 0.9567087254
$$
With $\tau = \frac{1}{2} + i\frac{\sqrt3}{6}$ we get
$$
f(e^{-\pi/\sqrt3}) = 2^{1/3} e^{-\pi\sqrt3/72} \approx 1.168211663
$$
With $\tau = i\frac{\sqrt{10}}{2}$ we get
$$
f(-e^{-\pi\sqrt{10}}) = 2^{1/4}\;
\left(\frac{1+\sqrt{5}}{2}\right)^{1/2}\;e^{\pi\sqrt{10}/24}
\approx 0.9999515311
$$
