1. ## Strange limit question

Hi All,

I want to know whether there is an easy way to see how fast

$\lim_{n \to \infty}(1 - (1-1/n^a)^n)$

goes to zero as a function of $a~(a > 1)$.

I already know (thanks Maple!) that for $1 \le a<2$ the function goes slower than $n^{-1}$, i.e.

$\lim_{n \to \infty}n (1 - (1-1/n^a)^n) = \infty~\text{for}~1 \le a < 2$

whereas

$\lim_{n \to \infty}n (1 - (1-1/n^a)^n) = 0~\text{for}~a>2$.

(and for $a=2$ the limit is 1). But is there a clear and easy way to see this? I can't even do the limit by hand because it seems to require an infinite number of applications of L'hopitals rule...

Note: Obviously the case $a=1$ is easy since I can substitute $e^{-1}$ and the thing goes to a constant .. but what about the cases when it really goes to zero? Obviously a=2 is special.

2. Ah, OK, nevermind, I think I see it using the binomial formula on the $(1 - (1-1/n^a)^n)$ term, giving a sum where one can deduce the stated property by factorization (the 1 and the first term in the series cancel).
3. You might also note that $\lim_{n\to\infty} (1+ \frac{1}{n})^n$ is commonly used as a definition of e. If we replace the "n" in $(1- \frac{1}{n^a})^n$ with $n^{1/a}$ (assuming, for the moment, that $a\ne 0$), we get $(1- \frac{1}{n})^{n/a}$. Now replace that n by "-n" to get $(1+ \frac{1}{n})^{-n/a}= [(1+ \frac{1}{n})^n]^{-1/a}= e^{-1/a}= \frac{1}{e^{1/a}}$