I know this should be easy, but I just can't find the proper search result. Thanks.

$\left(1-\frac1n\right)^n$, what is the estimation value when $n$ is very large?

Some follow-up,

If $n = 100$, what is the formula to calculate this?

  • $\begingroup$ Define "estimation". I know that for large, large $\;n\;$ a pretty good estimation is $\;e^{-1}\;$ ... $\endgroup$ – DonAntonio Sep 18 '13 at 23:55
  • $\begingroup$ I've never heard of an "estimation value" but $\lim_{n \to \infty} (1- 1/n)^n = 1/e$, so $(1- 1/n)^n$ is close to $1/e$ if $n$ is large and positive. This is a standard calculus exercise. $\endgroup$ – Stefan Smith Sep 18 '13 at 23:59

The estimation value when n is very large is $1/e$, where $e$ is the well known mathematical constant.

\begin{gather} \lim\limits_{n\to\infty}\left(1-\dfrac{1}{n}\right)^{-n}=\lim\limits_{n\to\infty}\left(\dfrac{n-1}{n}\right)^{-n}= \\ =\lim\limits_{n\to\infty}\left(\dfrac{n}{n-1}\right)^{n}=\lim\limits_{n\to\infty}\left[\left(1+\dfrac{1}{n-1}\right)^{n-1}\left(1+\dfrac{1}{n-1}\right)\right]=e \end{gather}

Therefore, what you are seeking is $e^{-1}$, or $1/e$.

  • 1
    $\begingroup$ Thanks for your reply. What if n = 100, can you give some link or formula on this? $\endgroup$ – melodrama Sep 19 '13 at 0:02
  • $\begingroup$ If n = 100, then the value is approximately 0.36603234127322950493061602657251738618971207663892369140595737269931704475072474818719654351002695040066156910065284327471823569680179941585710535449170757427389035006098270837114978219916760849490001. Source: www.wolframalpha.com $\endgroup$ – user85362 Sep 19 '13 at 0:07

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