What's the limit of the sequence \lim_{n\to\infty}\frac{n!}{n^n}

Shirley Thompson

Shirley Thompson

Answered question

2022-01-16

Whats

Answer & Explanation

Robert Pina

Robert Pina

Beginner2022-01-16Added 42 answers

You can prove that
n!<(n+12)n
Now observe that
0limnn!nn<limn(n+12)nnn=limn12n(n+1)nnn
We know that 12n0 as n. If you know that [n+1n]ne as n, then youre
Natalie Yamamoto

Natalie Yamamoto

Beginner2022-01-17Added 22 answers

No , both are equal in terms of their order of growth, obviously in terms of values nn is greater.
nn(n1)(n2)(n3)
Here , n multiplies by n times
So, nnn - something ,
here definitely something is less than nn
And in a function generally we considers highest order function among others , so nnn in terms of order of growth. It's simply like f(n)=n2+n+2, g(n)=n2+2 , here both are qudartic as well as they have n2 as highest order function , so they are equal in growth.
alenahelenash

alenahelenash

Expert2022-01-24Added 556 answers

Let un=n!nn then un+1=(n+1)!(n+1)(n+1). Now L=limnun+1unlimn(n+1)!(n+1)(n+1)×nnn! =limn(nn+1)n=e1<1 By ratio test for sequences the limit limnn!nn=0

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?