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

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: $$0 \leq \bar L \leq \infty$$

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This is related by integration by parts to this question of mine. –  Sasha Nov 22 '11 at 17:06
@Sasha: thanks! I think those answers will work fine (the monthly's formulation is very easy to adjust). too bad the suggest a question feature doesn't know integration by parts! –  Jack Schmidt Nov 22 '11 at 17:23
52*square root of pie=513.2194289 –  user48815 Nov 9 '12 at 18:03

2 Answers 2

up vote 2 down vote accepted

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.

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thanks, this looks good. I think this exactly fills in the red step in Srivatsan's answer: instead of using Stirling's approximation as a black box, use Laplace's method to get an appropriate estimate for our slightly smaller integral. –  Jack Schmidt Nov 22 '11 at 18:20
@Zarrax Where can I find theory on this method? I'm aware for the asymptotics of the exponent but this seems a more rigorous method. –  Pedro Tamaroff Feb 19 '12 at 23:18
@Peter's_method –  Zarrax Feb 20 '12 at 15:10
@Zarrax Thank you. Is there a non-wikipedia source I can use, a book for example? (No irony intended. An online paper or a .pdf article'd work as well.) –  Pedro Tamaroff Feb 20 '12 at 15:12
@Peter I don't know offhand but you should be able to find something by googling around. It's very well-known. –  Zarrax Feb 20 '12 at 15:15

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.

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Thanks, this looks good, and it gives a more or less effective upper bound as well. I'll wait until end of day my time to accept in case someone can tighten up your step in red (which surprisingly didn't make all that much difference between your bound and the actual value). –  Jack Schmidt Nov 22 '11 at 16:40

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