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!