# Prove: $\lim\limits_{n \to \infty} \int_{0}^{\sqrt n}(1-\frac{x^2}{n})^ndx=\int_{0}^{\infty} e^{-x^2}dx$

I'd like your help with the following claim to prove:

$$\lim_{n \to \infty} \int_{0}^{\sqrt n}\left(1-\frac{x^2}{n}\right)^ndx=\int_{0}^{\infty} e^{-x^2}dx.$$

I think I should use the claim:

Let $a,b$ two real numbers and $\{f_n\}$ a sequence of continuous functions on $\left[a,b\right]$ which converges uniformly to $f$ on $[a,b]$. Then $$\lim_{n\to\infty}\int_a^bf_n(t)dt=\int_a^bf(t)dt.$$

But for this, I must prove that $(1-\frac{x^2}{n})^n$ uniformly converges to $e^{-x^2}$. How do I do that? One of the hardest and trickiest things to do is to prove this uniformly converges.. every function has its own way. I believe that the more examples I'll see the easiest it will be. Thanks again!

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Apply Dini's theorem to the even and odd subsequences to get uniform convergence ( en.wikipedia.org/wiki/Dini%27s_theorem ) – Listing Dec 14 '11 at 9:03
You know the limit $\lim\limits_{n\to\infty}\left(1+\frac{z}{n}\right)^n=\exp\,z$, no? – J. M. Dec 14 '11 at 9:09
Yeah, I know the limit and indeed Dini will be great here. Thanks! – Jozef Dec 14 '11 at 11:41

Although perhaps not the best approach here, I give a proof with bare hands. The idea is to bound the integrand $f(x) = \left( 1- \frac{x^2}{n} \right)^n$, and hence the integral, directly.

• Upper bound. Since $1 - z \leqslant \mathrm e^{-z}$ for all $z$, we have we have $f(x) \leqslant \mathrm e^{-x^2}$ for $0 \leqslant x \leqslant \sqrt{n}$. Therefore, $$\int_{0}^{\sqrt{n}} f(x) \, dx \leqslant \int_{0}^{\sqrt{n}} \mathrm e^{-x^2} \, dx \lt \int_{0}^{\infty} \mathrm e^{-x^2} \, dx.$$

• Lower bound. Here we need the standard estimate $1 -z \geqslant \mathrm e^{-z - z^2}$ valid for $0 \leqslant z \leqslant \frac14$. (This can be obtained by using Taylor's theorem for $\log (1-z)$ at $z=0$.) Therefore, it follows that for $0 \leqslant x \leqslant \frac12 \sqrt{n}$, $$f(x) \geqslant \exp \left( - x^2 - \frac{x^4}{n} \right) .$$ In particular, for $0 \leqslant x \leqslant n^{1/8} \lt \frac12 \sqrt{n}$ (for large enough $n$), we have $f(x) \geqslant \mathrm{e}^{-x^2} \mathrm e^{- \frac{1}{\sqrt{n}}}$.

Therefore, \begin{align*} \int_0^{\sqrt{n}} f(x) \, dx &\geqslant \int_0^{n^{1/8}} f(x) \, dx \\ &\geqslant \int_0^{n^{1/8}} \mathrm{e}^{-x^2} \mathrm e^{- \frac{1}{\sqrt{n}}} \, dx \\ &= \mathrm e^{-\frac{1}{\sqrt{n}}} \cdot \int_0^{n^{1/8}} \mathrm{e}^{-x^2}\, dx \end{align*} Notice that the sequence $\mathrm e^{-\frac{1}{\sqrt{n}}} \cdot \int_0^{n^{1/8}} \mathrm{e}^{-x^2}\, dx$ converges to $\int_0^\infty \mathrm e^{-x^2} \, d x$.

Therefore by the sandwich theorem, we can conclude that $$\int_0^{\sqrt{n}} f(x) \, dx \to \int_0^\infty \mathrm e^{-x^2} \, dx.$$

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Indeed, $1-z\ge e^{-z-z^2}$ is valid on $[0,1/2]$, at least. First, consider where the derivative of $z+z^2+\log(1-z)$ is $0$: $$1+2z-\frac{1}{1-z}=0\Leftrightarrow z-2z^2=0\Leftrightarrow z\in\{0,1/2\}$$ Thus, $z+z^2+\log(1-z)$ is monotonic on $[0,1/2]$ and non-negative at both ends. Thus, $\log(1-z)\ge-z-z^2$ and therefore, $1-z\ge e^{-z-z^2}$ – robjohn Dec 15 '11 at 8:25

I fail to see how the behaviour of the integrals on $[a,b]$ could yield the $[0,+\infty)$ case.

Hint for a solution: consider the functions $f$ and $f_n$ defined on $\mathbb R_+$ by $f(x)=\mathrm e^{-x^2}$ and $$f_n(x)=\left(1-\frac{x^2}n\right)^n\cdot[x\leqslant\sqrt{n}].$$ Show that the sequence $(f_n)_{n\geqslant1}$ is increasing and that $f_n\to f$ pointwise. Then, find a theorem in your notes which guarantees that, in this setting, the integral of $f_n$ converges to the integral of $f$.

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+1, but I just wanted to add: I think this might be homework for an analysis course before measure theory, and the OP might not be allowed to use DCT or MCT. (Unless you were referring to something else?) However in that case he probably should of tagged it as homework so we wouldn't have to struggle trying to read his mind... – Eric Naslund Dec 14 '11 at 9:37
@Eric, I agree with your comment, in fact I considered asking the results known to the OP... As you say, Lebesgue would be a killer here. There are also convergence theorems for generalized Riemann integrals (which explains the last sentence of my post). – Did Dec 14 '11 at 10:11

If you want to use uniform convergence, you can do the following. Since $\ln(1 - y) \leq -y$ for $0 \leq y < 1$ (look at the power series), you have $n\ln(1 - {x^2 \over n}) \leq -n{x^2 \over n} = -x^2$, and taking exponentials you get $(1 - {x^2 \over n})^n \leq e^{-x^2}$ on your range of integration.

So you can fix some $a$ and divide your integral into $0$ to $a$ and $a$ to $\sqrt{n}$ portions. The second portion will be bounded by $\int_a^{\infty}e^{-x^2}\,dx$ for all $n$ by the above. On the first portion, since $n\ln(1- {x^2 \over n}) = -x^2 + O({x^4 \over n})$, the function $n\ln(1- {x^2 \over n})$ converges uniformly to $-x^2$. Taking exponentials $(1 - {x^2 \over n})^n$ converges to $e^{-x^2}$ uniformly on $0 \leq x \leq a$. So you can apply the uniform convergence theorem on this first part. The second integral is never more than $\int_a^{\infty}e^{-x^2}\,dx$ by the above. So letting $a$ go to infinity gets you the result.

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