Limit of the integral is the square root of pi over 2

Question

$$\bar L = \displaystyle \limsup_{n\to\infty} \frac{1}{\sqrt{n}} \int_0^\infty e^{-x}\left(1+\frac{x}{n}\right)^n ~dx$$

How do you show the limit superior is finite?

I actually am relatively certain the limit itself exists:

$$L = \displaystyle \lim_{n\to\infty} \frac{1}{\sqrt{n}} \int_0^\infty e^{-x}\left(1+\frac{x}{n}\right)^n ~dx \stackrel{?}{=} \sqrt{\frac{\pi}{2}}$$

I've tried the tricks I know, and nothing has worked out.

Of course if we interchange the limsup and the integral, then we get the integrand to be $e^{-x} e^x = 1$, and so the integral itself is ∞, and if we bring the square root inside first, and then interchange the limsup and integral we get the integrand to be 0, so I am faily confident of the following: $$0 \leq \bar L \leq \infty$$

Answer 1

I can show that the sequence is bounded, though the existence of the limit does not follow from this approach.

Make the substitution $t = n + x$. Then $$ \begin{align*} \frac{1}{\sqrt{n}} \int_0^\infty e^{-x}\left(1+\frac{x}{n}\right)^n ~dx &= \frac{1}{n^n \sqrt{n}} \int_0^\infty e^{-x} (n+x)^n ~dx \\ &= \frac{1}{n^n \sqrt{n}} \int_n^\infty e^{-(t - n)} t^n ~dt \\ &= \frac{e^n}{n^n \sqrt{n}} \int_{\color{Red}{n}}^\infty e^{-t} t^n ~dt \\ &\leqslant \frac{e^n}{n^n \sqrt{n}} \int_{\color{Red}{0}}^\infty e^{-t} t^n ~dt \\ &\stackrel{(a)}{=} \frac{e^n}{n^n \sqrt{n}} n! \\ &\stackrel{(b)}{\leqslant} \sqrt{2 \pi} ( 1 + o(1) ), \end{align*} $$ where we used (a) a standard identity (see e.g., Intuition for the definition of the Gamma function?), and (b) Stirling's formula.

Answer 2

First notice that by changing variables $x = ny$ you have $$\int_0^{\infty}e^{-x}(1 + {x \over n})^n\,dx = n\int_0^{\infty}e^{-ny}(1 + y)^n\,dy $$ $$= n\int_0^{\infty}e^{n(\ln(1 + y) - y)}\,dy $$ The idea is to use Laplace's method for evaluating integrals of $e^{Mf(x)}$. The phase $\phi(y) = \ln(1 + y) - y$ satisfies $\phi'(0) = 0$ and $\phi'(x)$ is nonzero elsewhere. Also note that $\phi''(0) = -1$. So asymptotically the integral will behave as $$\int_0^{\infty}e^{-{ny^2 \over 2}}\,dy$$ To be rigorous, Laplace's method gives that for a small $\epsilon > 0$ asymptotically you have $$\int_{-\epsilon}^{\infty}e^{n(\ln(1 + y) - y)}\,dy \sim e^{n\phi(0)}\sqrt{2\pi \over n|\phi''(0)|}$$ The right hand side is $\sqrt{2\pi \over n}$ here. Because you're starting at $y = 0$ here, you take half of this, or $\sqrt{\pi \over 2} n^{-{1 \over 2}}$. Then multiply by the $n$ above and you get $\sqrt{\pi \over 2} n^{{1 \over 2}}$, the desired asymptotics.