# Calculating $\lim\limits_{n\to\infty}\frac{n!\cdot e^n}{n^n}$

by 63677   Last Updated October 10, 2019 03:20 AM - source

I tried using the same trick as $$\lim\limits_{n\to\infty}\frac{n!}{n^n}$$, where you compare the terms one to one.

$$(\frac{1}{n})(\frac{2}{n})(\frac{3}{n})...(\frac{n}{n})\cdot e^n$$ = $$(\frac{e}{n})(\frac{2e}{n})(\frac{3e}{n})\cdots(\frac{ne}{n})$$

I can't really figure out what to do after this. Some of the terms are less than one and some are greater than one. It would be great if someone could lead me in the right direction of what to do here.

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Your are going in right direction See as n goes to infinity n>>> e so the first term becomes very less.now for the last term n would cancel out itself so last term is e

So limit equals to 0 because maximum value that is multiplied is e and every other value is very close to zero

Harshit Gupta
October 10, 2019 03:00 AM

Use Sterling formula and you have $$\frac {n! e^n}{n^n}\approx \sqrt {2\pi n}(n/e)^n (e/n)^n = \sqrt{2\pi n} \to \infty$$

October 10, 2019 03:04 AM

Apply Stirling's approximation so that as $$n\to\infty$$

$$n! \sim \left(\frac{n}{e}\right)^{n}\sqrt{2\pi n}$$

therefore

$$\frac{n!\cdot e^n}{n^n}\sim \sqrt{2\pi n}$$

from which $$\sqrt{2\pi n}$$ increases monotonically as $$n\to\infty$$

$$\lim\limits_{n\to\infty}\sqrt{2\pi n}=\infty$$

thus

$$\lim\limits_{n\to\infty}\frac{n!\cdot e^n}{n^n}=\infty$$

Axion004
October 10, 2019 03:07 AM

## Trigonometric Limit 0*infinity

Updated November 10, 2017 20:20 PM