Proving that $e^{n\log(1-\frac{x^2}{n})}$ is monotonic in $n$

Question

I am trying to prove that the sequence of functions $f_n = (1-\frac{x^2}{n})^n$ uniformly converges to $f = e^{-x^2}$, in any closed interval $[-A,A]$, for some $A>0$.

I want to use dini's theorem but for that I need to prove that for every $x \in [-A,A]$, the sequence is monotonic in $n$.

I thought about proving that for every $x \in [-A,A]$, and for every $n > A^2$, the sequence $f_n(x) = e^{n\log(1-\frac{x^2}{n})}$ is monotonically increasing, but I didn't succeed.

Answer 1

Note that $e^{n\log\left(1-\frac{x^2}{n}\right)}=\left(1-\frac{x^2}{n}\right)^n$. Then, letting $a_n=\left(1-\frac{x^2}{n}\right)^n$, we have for $n>x^2$

$$\begin{align} \frac{a_{n+1}}{a_n}&=\frac{\left(1-\frac{x^2}{n+1}\right)^{n+1}}{\left(1-\frac{x^2}{n}\right)^n}\\\\ &=\left(1-\frac{x^2}{n}\right)\left(1+\frac{x^2}{(n+1)(n-x^2)}\right)^{n+1}\tag 1\\\\ &\ge \left(1-\frac{x^2}{n}\right)\left(1+\frac{x^2}{n-x^2}\right)\tag2\\\\ &=1 \end{align}$$

where in going from $(1)$ to $(2)$ we used Bernoulli's Inequality.

Inasmuch as $\frac{a_{n+1}}{a_n}\ge1$, $a_n$ is increasing. And we are done!

Answer 2

Let $F(x,y)=\left(1+\frac xy\right)^y$. Then$$\frac{\partial F}{\partial y}(x,y)=\left(1+\frac xy\right)^y\cdot\frac{-x+(x+y)\log\left(1+\frac xy\right)}{x+y}.$$Notice that$$-x+(x+y)\log\left(1+\frac xy\right)>0\iff \log\left(1+\frac xy\right)>\frac x{x+y}=\frac{\frac xy}{1+\frac xy}.$$It is easy to prove that$$(\forall x>-1):\log(1+x)>\frac x{1+x}.$$Therefore, $F$ increases when $y$ does. You deduce from this that the sequence $\left(\bigl(1+\frac xn\bigr)^n\right)_{n\in\mathbb N}$ is an increasing one. Putting $-x^2$ instead of $x$, you get what you want.

Answer 3

Binomial theorem comes to the rescue. If $a_{n} $ is your sequence then $b_{n} =1/a_{n}=(1-x^{2}/n)^{-n}$. We can easily show that if $n>x^{2}$ then $b_{n}>0$ and $b_{n}$ is decreasing. Also to simplify typing let's use $y$ in place of $x^{2}$ in what follows.

Via binomial theorem we have for $n>y$ $$b_{n} =\left(1-\frac{y}{n}\right)^{-n}=1+y +\dfrac{1+\dfrac{1}{n}}{2!}y^{2}+\dfrac{\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{2}{n}\right)}{3!}y^{3}+\dots$$ We can see that as $n$ increases each term decreases. Thus $b_{n} $ is decreasing and $a_{n} =1/b_{n}$ is increasing.