TITLE: How to find the limit that involves the factorials?
QUESTION [7 upvotes]: $$ \lim_{n\to\infty}\left(\int_a^b((x-a)(b-x))^n\mathrm{d}x\right)^{\frac{1}{n}}$$
I have found the form of the integrand to be $$\frac{(n!)^2}{(2n+1)!}(b-a)^{2n+1}$$ Now, splitting the limit into two different parts i would need to solve $$\lim_{n\to\infty}(\frac{(n!)^2}{(2n+1)!} )^{\frac{1}{n}}$$ My mind went to Stirling's approximation formula but i don't think it could help.

REPLY [2 votes]: $$y=\left(\frac{(n!)^2}{(2n+1)!} (b-a)^{2n+1}\right)^{\frac{1}{n}}$$
$$\log(y)=\frac{1}{n}\Bigg[2\log(n!)+(2n+1)\log(b-a)-\log((2n+1)!) \Bigg]$$ Using Stirling approximation and continuing with Taylor expansions
$$n\,\log(y)=\log\Big[\frac 14{(b-a)^2} \Big]n+\frac{1}{2} \log \left(\frac{\pi  (b-a)^2}{4 n}\right)-\frac 3 {8n}+O\left(\frac{1}{n^2}\right)$$
$$\log(y)=\log\Big[\frac 14{(b-a)^2} \Big]+\frac{1}{2n} \log \left(\frac{\pi  (b-a)^2}{4 n}\right)+O\left(\frac{1}{n^2}\right)$$
$$y=e^{\log(y)}\sim\frac 14{(b-a)^2}\exp\Bigg[\frac{1}{2n} \log \left(\frac{\pi  (b-a)^2}{4 n}\right) \Bigg]$$