Show $\lim_{n \to \infty} \sup_{x \in [0,1]} (e^{-nx} - (1-x)^n) = 0$?

Question

I am trying to show that $$\lim_{n \to \infty} \sup_{x \in [0,1]} (e^{-nx} - (1-x)^n)=0.$$ I have tried expanding $e^{-nx}$ with its Taylor series and $(1-x)^n$ with Newton's binomial theorem. The resulting Taylor series is $\sum_{k=2}^{\infty} (-1)^k \left(\frac{n^k}{k!} - {n \choose k}\right)x^k.$

I also tried taking the first derivative with respect to $x$ to find the point in $(0,1)$ where $e^{-nx} - (1-x)^n$ is maximized (in $x$ as a function of $n$), but I was only able to find the hard-to-solve $\frac{n}{1-n} = \frac{\ln(1-x)}{x}.$

Answer 1

Let $f_n(x) =e^{-nx} - (1-x)^n $.

I will show that for $0 < x < 1$, $f_n(x) \ge 0$ and if $x > \dfrac1{\sqrt{n}}$ then $f_n(x) \lt e^{-\sqrt{n}} $. and if $0 < x < c_0 \approx 0.683803 $ (the root of $\ln(1-x)+x+x^2 =0$) then $f_n(x) < x $.

Therefore $0 < f_n(x) \le\dfrac1{\sqrt{n}} $.

For $0 < x < 1$, $e^x < \dfrac1{1-x} $ (compare the power series) so $e^{-x} > 1-x $ so $e^{-nx} > (1-x)^n $.

Therefore $f_n(x) > 0$.

If $x > \dfrac1{\sqrt{n}}$ then $e^{-nx} \lt e^{-\sqrt{n}} $ so $f_n(x) \lt e^{-\sqrt{n}} $.

$\begin{array}\\ f_n(x) &=e^{-nx}-(1-x)^n\\ &=e^{-nx}-e^{n\ln(1-x)}\\ &=e^{-nx}(1-e^{nx+n\ln(1-x)})\\ &<e^{-nx}(1-e^{nx+n(-x-x^2)}) \qquad 0 < x < c_0\ (*)\\ &=e^{-nx}(1-e^{-nx^2})\\ &<e^{-nx}(1-(1-nx^2))\\ &=nx^2e^{-nx}\\ &\le x \qquad\text{since } z e^{-z} \le 1\\ \end{array} $

$(*) \ \ln(1-x) \gt -x-x^2 $ for $0 < x < c_0 \approx 0.683803 $.

Answer 2

To get rid of the $e^{-n x}$ term, use the formula that you just found: $\frac{n}{1-n} = \frac{\log(1-x)}{x}$ (just move the $x$ in the denominator of the RHS to the LHS). That is going to give you a formula of the form $(1-x)^{n-1} - (1-x)^n$. From here, the result should be almost immediately.

Answer 3

actually if you focus on the limit $$\lim_{n \to \infty}\displaystyle e^{\displaystyle-nc} -(1-c)^n$$ where c is a real number between $0$ and $1$then if $c \neq 0$ then the exponential term goes to zero and the other bracket also and if $c = 0$ the value of the limit is zero so this can be proved directly without evaluating the maximum

Answer 4

Use the result: $$\color{red}{0\le e^{-nx}-(1-x)^n\le nx^2 e^{-nx} \qquad (0\le x\le 1),}$$ which can be refered in key inequality. When $n>2$, $f_n(x)=nx^2 e^{-nx}$ has maximum at $\frac{2}{n}\in(0,1)$. So $$\sup_{x\in[0,1]}\left(e^{-nx}-(1-x)^n\right)\leq\frac{4}{n}e^{-2}\to 0.$$