You can prove that n!&lt;(n+12)n Now observe that 0≤limn→∞n!nn&lt;limn→∞(n+12)nnn=limn→∞12n⋅(n+1)nnn We know that 12n→0 as n→∞. If you know that [n+1n]n→e as n→∞, then youre

Let un=n!nn then un+1=(n+1)!(n+1)(n+1). Now L=limn→∞un+1un⇒limn→∞(n+1)!(n+1)(n+1)×nnn! =limn→∞(nn+1)n=e−1&lt;1 By ratio test for sequences the limit limn→∞n!nn=0

You can prove that n!&lt;(n+12)n Now observe that 0≤limn→∞n!nn&lt;limn→∞(n+12)nnn=limn→∞12n⋅(n+1)nnn We know that 12n→0 as n→∞. If you know that [n+1n]n→e as n→∞, then youre

Let un=n!nn then un+1=(n+1)!(n+1)(n+1). Now L=limn→∞un+1un⇒limn→∞(n+1)!(n+1)(n+1)×nnn! =limn→∞(nn+1)n=e−1&lt;1 By ratio test for sequences the limit limn→∞n!nn=0

Question: "Whats" - Plainmath

Shirley Thompson

Answered question

2022-01-16

Whats

Answer & Explanation

Robert Pina

Beginner2022-01-16Added 42 answers

You can prove that

n! < {(\frac{n + 1}{2})}^{n}

Now observe that

0 \leq lim_{n \to \infty} \frac{n!}{n^{n}} < lim_{n \to \infty} \frac{{(\frac{n + 1}{2})}^{n}}{n^{n}} = lim_{n \to \infty} \frac{1}{2^{n}} \cdot \frac{{(n + 1)}^{n}}{n^{n}}

We know that

\frac{1}{2^{n}} \to 0

n \to \infty

. If you know that

{[\frac{n + 1}{n}]}^{n} \to e

n \to \infty

, then youre

Natalie Yamamoto

Beginner2022-01-17Added 22 answers

No , both are equal in terms of their order of growth, obviously in terms of values

n^{n}

is greater.

n \neq n (n - 1) (n - 2) (n - 3)

Here , n multiplies by n times
So,

n \neq n^{n}

- something ,
here definitely something is less than

n^{n}

And in a function generally we considers highest order function among others , so

n \neq n^{n}

in terms of order of growth. It's simply like

f (n) = n^{2} + n + 2, g (n) = n^{2} + 2

, here both are qudartic as well as they have

n^{2}

as highest order function , so they are equal in growth.

alenahelenash

Expert2022-01-24Added 556 answers

Let

u_{n} = \frac{n!}{n^{n}}

then

u_{n + 1} = \frac{(n + 1)!}{(n + 1)^{(n + 1)}} .

Now

L = lim_{n \to \infty} \frac{u_{n + 1}}{u_{n}} \Rightarrow lim_{n \to \infty} \frac{(n + 1)!}{(n + 1)^{(n + 1)}} \times \frac{n^{n}}{n!}

= lim_{n \to \infty} (\frac{n}{n + 1})^{n} = e^{- 1} < 1

By ratio test for sequences the limit

lim_{n \to \infty} \frac{n!}{n^{n}} = 0

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