How to maximize $\prod_{j=0}^{l-1} \frac{2n-2-j}{n+l-1-j}$ with respect to $l$ fixed $n$ I am trying to prove the convergence of a series and I have a coefficient of the form
$$\prod_{j=0}^{l-1} \frac{2n-2-j}{n+l-1-j}=\frac{(2n-2)!\,(n-1)!}{(2n-2-l)!(n-1+l)!}$$
with $0\le l \le 2n-2$. Given $n$ I want to find the upper bound of this.
I figured out that I need to search for the maximum in $0\le l \le n-1$. So I am looking for a closed form expression of
$$\max_{l\in\{0,\dots,n-1\}}\prod_{j=0}^{l-1} \frac{2n-2-j}{n+l-1-j}$$
However I am stuck with this. The fact that $l$ is also in the limit of the product confuses me a lot.
Is there any straightforward way to calculate this?

The expression in terms of factorials is much more convenient than the expression in terms of the product.  In particular, note that the arguments of the factorials in both the numerator and the denominator sum to $3n-3$; this means that you can write the ratio as a ratio of binomial coefficients by multiplying by $\frac{(3n-3)!}{(3n-3)!}$.  Once you do that, factor out the 'constant' ($n$-dependent) term, look at the term dependent on $l$, and use the 'bitonicity' of the binomial coefficient that $a\choose b$ is always largest for $b=\lfloor\frac a2\rfloor$.  Your ultimate bound will involve terms of the form $2c\choose c$ and $3d\choose 2d$, for $c$ and $d$ that wind up being some specific functions of $n$; Stirling's formula will then handle these forms nicely.
