Question

Compute.\[\lim_{n\to\infty} \left(\frac{n^n}{n!}\right)^{1/n}\]

Answer 1

\[\lim_{n\to \infty}\frac n{(n!)^{1/n}}\]Can I do this?

Answer 2

inf^0 on the bottom

Answer 3

\[\large \lim_{n\to \infty}e^{\ln n - \frac1n\ln n!}\]

Answer 4

i was thinking the same.

Answer 5

I've got a bad feeling about this .... does it converge??

Answer 6

But I don't know how to get it from here :/
We can't even use l'hospital.

Answer 7

ln n! = ln n + ln n-1 + ln n-2 + ....

Answer 8

The answer is e, I think.

Answer 9

Yes ... it converges

Answer 10

ans 0

Answer 11

No ... i think i forgot to multiply by n.

Answer 12

let's ask wolf bro first.

Answer 13

wolf bro says e

Answer 14

\[\frac n{(n!)^{1/n}} = n\cdot\frac{1}{n^{1/n}}\cdot \frac{1}{(n-1)^{1/n}}\cdots \frac{1}{2^{1/n}}\cdot\frac{1}{1^{1/n}}\]

Answer 15

Hmm or maybe
\[\left(\frac{n}{n-1}\cdot \frac n {n-2}\cdots\frac n 2 \cdot\frac n1\right)^{1/n}\]

Answer 16

yes ... that might be

Answer 17

let's put it in ln on the top of e

Answer 18

\[\huge e^{\frac1n \ln\left(\frac{n}{n-1}\cdot \frac n {n-2}\cdots\frac n 2 \cdot\frac n1\right)}\]

Answer 19

I wish I knew more math other than the fancy tex.

Answer 20

expand ln and ln(n/n) + ln(n/n-1) ...

Answer 21

and not n-1 it's n at the beginning.

Answer 22

But we can cancel, can't we? ln1 = 0

Answer 23

yes .. only first one.

Answer 24

1/n(ln n/(n-1) + ln n/(n-2) + ... + ln (n/1))

Answer 25

1/n(ln n/(n) + ln n/(n-1) + ... )

Answer 26

But ln n -> infity, maybe we can do something with that 1/n

Answer 27

(ln n - ln(n-1))/n + (ln n - ln(n-2))/n + ...

Answer 28

It's the same as before

Answer 29

Let's use L'hospital rule here 1/n + (ln n - ln(n-1))/n + (ln n - ln(n-2))/n + ...

Answer 30

(ln n - ln(n-1))/n = (1/n - 1/(n-1)) / 1 + (1/n - 1/(n-2)) + ...

Answer 31

stuck ... where does out one come from??

Answer 32

Oh hmm Nice
1/n - 1/(n-1) + 1/n - 1/(n-2) + ... + 1/n - 1
1 - (1/(n-1) + 1/(n-2) + ... + 1/(2) + 1)

Answer 33

-(1/(n-1) + 1/(n-2) + ... 1/2)

Answer 34

Oh wait

Answer 35

hmm i feel i did something wrong

Answer 36

maybe @eliassaab can help.

Answer 37

i am out of clue.

Answer 38

I will try searching on mathstackexchange someone must have asked this question.

Answer 39

that would spoil the fun.

Answer 40

LOL

Answer 41

I am getting answer zero.

Answer 42

yeah :/
hmm then i will bump the question.

Answer 43

e^0

Answer 44

wolf bro says this must be
http://www.wolframalpha.com/input/?i=lim+n-%3Einf+1%2Fn+ln%28n%5En%2Fn%21%29

Answer 45

\[\left(\frac{n}{n-1}\cdot \frac n {n-2}\cdots\frac n 2 \cdot\frac n1\right)^{1/n}= \left(\frac1{1 - 1/n} \cdot \frac1 {1-2/n}\cdots\right)^{1/n}\]

Answer 46

ln 1 = 0 ,,,, damn.

Answer 47

i think i understand the error of my method.

Answer 48

I am out of clues.

Answer 49

the left would be infinity.

Answer 50

\[\Large e^{\ln n - \frac{\ln n!}n} =\frac {e^{\ln n} }{e^{-\frac{\ln n!}n}}\]

Answer 51

Sorry. hehe

Answer 52

What am I doing!?!?!

Answer 53

the top would be infinity and the bottom would be zero.

Answer 54

Ideas generate like that.

Answer 55

\[\Large e^{\ln n - \frac{\ln n!}n} =\frac {e^{\ln n} }{e^{\frac{\ln n!}n}}\]Hmm are you sure bottom would be infinity n! >>> n

Answer 56

Can't we apply l'hospital on the bottom part now? maybe somehow we will get e^(ln n-1) hehe

Answer 57

I don't think we can :/

Answer 58

the bottom part woulbe be e^1

Answer 59

The problem is with n! at first it seems like a perfect variable but at the end it gets constant, and then you wonder what happens in between.

Answer 60

yeah ... that's why my method is not working. and n -something = n/some no

Answer 61

@phi @nikvist

Answer 62

@mr.math

Answer 63

(a) Show, using upper and lower Rieman Sums that
\[ \int_1^n \ln(x) \le \sum_{k=2}^n \ln (k) \le \int_1^n \ln(x) + \ln(n)

\]
(b) Compute
\[
\int_1^n \ln(x)
\]

Use (a) and (b) to prove
\[

(c) \quad e \left ( \frac n e \right)^n\le n! \le e \left ( \frac n e \right)^n n

\]
use (c)
to show that the sequence in question tends to e.

Answer 64

I think you can get the answer from third post ...

Answer 65

Hmm third post?

Answer 66

limn→∞ e^(lnn−1/n *lnn!)
substituting what you done before will give the answer.
Well I still like approximation way.
search for Srinivasa Ramanujan
http://en.wikipedia.org/wiki/Factorial

Answer 67

ohh hmm i will search for him, thanks.