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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 ( ) – 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
up vote 7 down vote accepted

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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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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