How to find the product $\Pi_{n=1 }^\infty (1+x^{2n-1 })$?

Question

$\Pi_{n=1 }^\infty (1+x^{2n-1 })$

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.

Answer 1

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).

Answer 2

$$ f(x) := \prod_{n=1 }^\infty (1+x^{2n-1 }),\qquad x \in \mathbb C, |x| < 1 $$ Converges to a holomorphic function. This is not an "elementary function". It can be written in terms of some known special functions.

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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 $$ f(-e^{-\pi\sqrt{10}}) = 2^{1/4}\; \left(\frac{1+\sqrt{5}}{2}\right)^{1/2}\;e^{\pi\sqrt{10}/24} \approx 0.9999515311 $$