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:
- Trying to prove that $\sum_{j=2}^\infty \prod_{k=1}^j \frac{2 k}{j+k-1} = \pi$
- On the closed form for $\sum_{m=0}^\infty \prod_{n=1}^m\frac{n}{4n-1}$
- Infinite Series $\sum\limits_{n=1}^{\infty}\frac{1}{\prod\limits_{k=1}^{m}(n+k)}$
- Closed form of series with factorial-squared denominator?
- Infinite series with a rising factorial?
