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Calculus Level 3

$\large \lim_{n \to \infty} \dfrac{n!}{!n}$

The limit above has a closed form. Find this closed form.

Give your answer to 3 decimal places.

Notations :

$!$ denotes the factorial notation. For example, $8! = 1\times2\times3\times\cdots\times8$ .
$!n$ denotes the number of derangements for $n$ different objects, $\displaystyle !n = n! \sum_{r=0}^n \dfrac{(-1)^r}{r!}$ .

The answer is 2.718.

3 solutions

Rishabh Jain
May 3, 2016

$\small{\bullet~!n=n!\left(\dfrac{1}{2!}-\dfrac{1}{3!}+\dfrac{1}{4!}+\cdots +(-1)^n\dfrac{1}{n!}\right)}$ $\bullet~$ And also note( by putting $x=-1$ in expansion of $e^x$ ):- $(\small{\color{#3D99F6}{e^{-1}=1-1+\dfrac{1}{2!}-\dfrac{1}{3!}+\dfrac{1}{4!}+\cdots}})$

The limit is therefore:-

$\displaystyle\lim_{n\to\infty}\dfrac{\cancel{n!}}{\cancel{n!}\small{\left(\color{#3D99F6}{\dfrac{1}{2!}-\dfrac{1}{3!}+\dfrac{1}{4!}+\cdots +(-1)^n\dfrac{1}{n!}}\right)}}$ $\Large =\dfrac{1}{e^{-1}}=e\approx\boxed{2.718}$

Very nice!

Geoff Pilling - 5 years, 1 month ago

Thanks.... $\ddot\smile$

Rishabh Jain - 5 years, 1 month ago

Abhay Tiwari
May 3, 2016

for $n>1$ the $!n$ can be written as $!n=[\frac{n!+1}{e}]$ , where $[]$ denotes the greatest integer (floor function), but since we are dealing with n which tends to infinity, so we will approach without the floor function.

$\lim_{n \to \infty} \dfrac{n!}{\frac{n!+1}{e}}$

$\lim_{n \to \infty} \dfrac{1}{\dfrac{1+\dfrac{1}{n!}}{e}}$

$\dfrac{1}{\frac{1+0}{e}}=\boxed{e}=\boxed{2.718}$

This is incorrect, when $n = 2$ , $!n = 1$ and $\dfrac{n!+1}e = \dfrac2e \approx 0.7358$ .

Pi Han Goh - 5 years, 1 month ago

Pi Han Goh :), if n=2, then $!n=[\frac{2+1}{e}]=1$ , actually it comes with a greatest integer(floor function[]) which I did not mention there as n was tending to infinity.