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