Help with a specific limit $\left( \dfrac{n-1}{n} \right)^n$ as $n \rightarrow \infty$

Allow me to preface this by saying this is not a homework problem. If I had had this thought four years ago when I was taking calculus, I probably could do it...

I'm trying to calculate the limit as $n \to \infty$ of $1-\left(\frac{n-1}{n}\right)^n$ - - it's a constant I'm inclined to call "natural chance of success". I have estimated this value to be ~0.632121 but would very much like to see how it could be calculated aside from the brute-force method i employed earlier.

The background of this would be... consider n = 2 (a coin). You are given two flips of the coin to get what you pick - what is the chance you'll get your chosen outcome, assuming of course it's a fair coin. The best way to go about this would be to say there's a 1/2 chance of failing, and you have 2 flips. This means you have $(1/2)^2$ chance of failure, being 1/4. 1-1/4 is 3/4, so your chance of success here is 3/4.

Now consider n = 6 (standard die). You get six rolls to get the number of your choice (assuming a fair die again). Again, you have a 5/6 chance to not get your choice, and 6 rolls at 5/6 chance would be $(5/6)^6$, or ~.334, giving you a ~.665 chance of success.

And I'm curious as n increases to infinity, what is your chance of success? Now again, I've estimated this with a double precision float (in Java) to be 0.63212 (actually, this was the point at which it could simply gain no more precision on the value, n = 296536) but this doesn't really give insight to the derivation of the number, merely its value.

So I'm hoping someone a little fresher on their integrals than I can help me out here.

Thanks!

Solution 1:

You are essentially trying to find the limit of $\left(1-\frac{1}{n}\right)^n$ as $n\to\infty$. Rewrite this as $e^{n\ln(1 - \frac{1}{n})}$. Since the exponential function is continuous, you have $$\lim_{n\to\infty} e^{n\ln(1-\frac{1}{n})} = e^{\lim_{n\to\infty}(n\ln(1-\frac{1}{n})}.$$

To compute that limit, rewrite and use L'Hopital's: $$\begin{array}{rcl} \lim_{n\to\infty}n\ln\left(1 - \frac{1}{n}\right) & = & \lim_{n\to\infty}\frac{\ln(1-\frac{1}{n})}{\frac{1}{n}}\\ & = & \lim_{n\to\infty} \frac{\left(\frac{1}{1-\frac{1}{n}}\right)\left(1-\frac{1}{n}\right)'}{(n^{-1})'}\\ & = & \lim_{n\to\infty}\frac{\quad\frac{n^{-2}}{1-\frac{1}{n}}\quad}{-n^{-2}}\\ & = & \lim_{n\to\infty}\frac{-1}{1-\frac{1}{n}} = -1. \end{array}$$ So, since you wanted $1$ minus this limit, your limit equals $1-e^{-1}$.

Solution 2:

If you know one of the definitions for the number $e$:

$$ e=\lim_{n\to\pm\infty}\Big(1+\frac{1}{n}\Big)^n, $$

then you could calculate your limit in the following way:

$$ \lim_{n\to\infty}\Big(1-\frac{1}{n}\Big)^n = \lim_{n\to\infty}\left[\Big(1+\frac{1}{-n}\Big)^{-n}\right]^{-1} = \frac{1}{e}. $$

Solution 3:

If you expand $\lim_{n\to\infty}\Big(1-\frac{1}{n}\Big)^n,$ by using the binomial formula, you get $1 - \frac{n}{n} + \frac{n*(n-1)}{2*n^2} - \frac{n*(n-1)*(n-2)}{6*n^3} + ...$ As n goes to infinity, this approaches $1 - \frac{n}{n} + \frac{n*n}{2*n^2} - \frac{n*n*n}{6*n^3} + ... $, which equals $1 - 1 + \frac{1}{2} - \frac{1}{6} + ...$ This is exactly the Taylor expansion of $e^x$ when $x=-1$

Surprisingly, wolfram alpha can actually match constants to approximations, like so: http://www.wolframalpha.com/input/?i=0.632121

That may be a resource that you could find use for in the future.

Solution 4:

Your question is motivated by probability, but in itself it's pure and elemental calculus:

$\lim_{x\to \infty} (1 - \frac{1}{x})^x = e ^{-1}$

From that, your number is 0.63212...