# Closed form for infinite series $\sum_{n=1}^\infty \prod_{j=1}^n \left[1-(\tfrac{j}{n}u+v)^2\right]^{-1} x^n$

I've been struggling to find a closed form for the following series that depends on $u$ and $v$ as parameters:

$$f(u,v,x) = \sum_{n=1}^\infty c_n x^n$$

with

$$c_n = \frac{1}{\prod_{j=1}^n \left[1-(\tfrac{j}{n}u+v)^2\right]} .$$

Any ideas or insight will be appreciated!

Here are some things I've tried. First, for the special case $u=0$, the series is a geometric series and the result is

$$f(x,0,v) = \frac{x}{1-x-v^2}.$$

In general, it's possible to factor the denominator in $c_n$ as follows

$$c_n = \frac{1}{\left(\prod_{j=1}^n \left[1+v+\tfrac{j}{n}u\right]\right) \left(\prod_{j=1}^n \left[1-v-\tfrac{j}{n}u\right]\right)} .$$

Not really sure if this factorization is useful, but Mathematica does something similar and immediately puts $c_n$ in the form

$$c_n = \frac{(-1)^n \left(\tfrac{n}{u}\right)^{(2n)}}{\left(1+\tfrac{(v+1)n}{u}\right)_n \left(1+\tfrac{(v-1)n}{u}\right)_n}$$ where $(\cdot)_n$ is a Pochhammer symbol (rising factorial). This form has some appealing symmetry, but Mathematica is unable to complete the sum.

Finally, here are some potentially-related problems: