How to prove $\lim_{n\to\infty}\frac{1}{\Gamma(n+1/2)}\int_{n}^{\infty} (\sqrt{t}-\sqrt{n}) \exp(-t)t^{n-1} dt = 0$?

Question

How to prove $$\lim_{n\to\infty}\frac{1}{\Gamma(n+1/2)}\int_{n}^{\infty} (\sqrt{t}-\sqrt{n}) \exp(-t)t^{n-1} dt = 0$$ ?

As I evaluated in MATLAB, the statement seems true.

Answer 1

Let $x=\chi(y)\geqslant 0$ be the solution of $x-\log(1+x)=y^2/2$ for $y\geqslant 0$. Then \begin{align*} J(n)&:=\int_n^\infty(\sqrt{t}-\sqrt{n})t^{n-1}e^{-t}\,dt \\\color{gray}{[t=n(1+x)]}\quad &=n^{n+1/2}e^{-n}\int_0^\infty\frac{\sqrt{1+x}-1}{1+x}\big((1+x)e^{-x}\big)^n dx \\\color{gray}{[x=\chi(y)]}\quad &=n^{n+1/2}e^{-n}\int_0^\infty\underbrace{\frac{\sqrt{1+\chi(y)}-1}{\chi(y)}}_{=\xi(y)}\,ye^{-ny^2/2}\,dy \\\color{gray}{[y=z/\sqrt{n}]}\quad &=n^{n-1/2}e^{-n}\int_0^\infty\xi(z/\sqrt{n})ze^{-z^2/2}\,dz. \end{align*} By DCT, the last integral tends to $\xi(0^+)\int_0^\infty ze^{-z^2/2}\,dz=1/2$ with $n\to+\infty$, hence $$\lim_{n\to+\infty}\frac{J(n)}{n^{n-1/2}e^{-n}}=\frac12.$$ It remains to apply Stirling's asymptotics for $\Gamma(n+1/2)$.