# Help with a details restriction $\left( \dfrac{n-1}{n} \right)^n$ as $n \rightarrow \infty$

Allow me to beginning this by claiming this is not a research trouble. If I had actually had this assumed 4 years earlier when I was taking calculus, I possibly can do it ...

I'm attempting to compute the restriction as $n \to \infty$ of $1-\left(\frac{n-1}{n}\right)^n$ - - it is a constant I'm likely to call "all-natural opportunity of success". I have actually approximated this value to be ~ 0.632121 yet would certainly significantly like to see just how maybe computed in addition to the brute - pressure method i used previously.

The history of this would certainly be ... take into consideration n = 2 (a coin). You are offered 2 turns of the coin to get what you select - what is the opportunity you'll get your picked end result, thinking certainly it is a reasonable coin. The most effective means to deal with this would certainly be to claim there is a 1/ 2 opportunity of falling short, and also you have 2 turns. This suggests you have $(1/2)^2$ opportunity of failing, being 1/ 4. 1 - 1/ 4 is 3/ 4, so your opportunity of success below is 3/ 4.

Currently take into consideration n = 6 (typical die). You get 6 rolls to get the variety of your selection (thinking a reasonable die once more). Once more, you have a 5/ 6 opportunity to not get your selection, and also 6 rolls at 5/ 6 opportunity would certainly be $(5/6)^6$, or ~.334, offering you a ~.665 opportunity of success.

And also I'm interested as n raises to infinity, what is your opportunity of success? Currently once more, I've approximated this with a double accuracy float (in Java) to be 0.63212 (in fact, this was the factor at which it can merely obtain say goodbye to accuracy on the value, n = 296536) yet this does not actually offer understanding to the derivation of the number, just its value.

So I'm wishing a person a little better on their integrals than I can aid me out below.

Many thanks!

You are basically searching for the restriction of $\left(1-\frac{1}{n}\right)^n$ as $n\to\infty$. Revise this as $e^{n\ln(1 - \frac{1}{n})}$. Given that the rapid function is continual, 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 calculate that restriction, revise and also make use of 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, given that you desired $1$ minus this restriction, your restriction amounts to $1-e^{-1}$.

If you recognize among the interpretations for the number $e$:

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

after that you can compute your restriction in the list below means:

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

If you expand $\lim_{n\to\infty}\Big(1-\frac{1}{n}\Big)^n,$ by utilizing 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 mosts likely to infinity, this comes close to $1 - \frac{n}{n} + \frac{n*n}{2*n^2} - \frac{n*n*n}{6*n^3} + ... $, which amounts to $1 - 1 + \frac{1}{2} - \frac{1}{6} + ...$ This is specifically the Taylor development of $e^x$ when $x=-1$

Surprisingly, wolfram alpha can in fact match constants to estimates, thus: http://www.wolframalpha.com/input/?i=0.632121

That might be a source that you can locate usage for in the future.

Related questions