Question
I was recently fiddling with some math that me wonder if there was nice asymptotic equation of the form $k<c$:
$$ \ln (kc-1)! \sim \ln(c!(c-1)! \dots(c-k)!) + c \cdot\alpha$$
where $c$ is a variable and $\alpha$ is a constant for large $c$. Can you prove the equation and determine the constants?
My Fiddling
Consider:
$$ \frac{(2c-1)!}{c!} = (c+1)(c+2)\dots(c+c-1)$$
Now also think of:
$$ (c-1)! = (c-1)(c-2)\dots (c-(c-1)) $$
Dividing both our equations:
$$ \frac{(2c-1)!}{c!(c-1)!} = \frac{(c+1)(c+2)\dots(c+c-1)}{(c-1)(c-2)\dots (c-(c-1))}$$
Taking $\ln$ both sides and multiplying $1/c$ both sides:
$$ \frac{1}{c}\ln \frac{(2c-1)!}{c!(c-1)!} = \frac{1}{c}\sum_{r=1}^{c-1} \ln \frac{1+\frac{r}{c}}{1-\frac{r}{c}} $$
Taking limit $c \to \infty$ and recognising the R.H.S as an integral:
$$ \lim_{c \to \infty}\frac{1}{c}\ln \frac{(2c-1)!}{c!(c-1)!} = \int_{0^+}^{1^-} \ln \Big( \frac{1+x}{1-x} \Big) dx$$
$$\int_{0^+}^{1^-} \ln \Big( \frac{1+x}{1-x} \Big) dx = (1-x)\ln(1-x) + (1+x)\ln(1+x) \Big|_{0^+}^{1^-} = 2 \ln 2$$
Hence:
$$ \ln(2c-1)! \sim \ln(c!(c-1)!)+ 2c \ln2 $$