Question

\[\lim_{n \rightarrow \infty}\frac{n}{\sqrt[n]{n!}}\]

Answer 1

woah hmm

\[\frac{1}{\sqrt[n]{\frac{n!}{n^n}}}\]

Answer 2

\[\large\frac{1}{\sqrt[n]{\frac{n(n-1)(n-2)\ldots (n-n)}{\underbrace{n.n.n.n.n.n.n...}_{\text{n times}}}}}\]

Answer 3

geometric mean of 1,2,...,n

Answer 4

\[ \frac{1}{\sqrt[n]{1(1-\frac{1}{n})(1 - \frac{2}{n})\ldots 1}}\]

Answer 5

\[\frac{1}{n} \to 0, n \to \infty\]

Answer 6

these things are usually 'e' :D

Answer 7

what is the answer btw?

i think i will have to consider the case when factorial is in the middle \[\frac{n-\frac{n}{2}}{n}\]hmm i get 1 - 1/2 here

is it zero? i don't think so i will have to redo it

Answer 8

hmmm, do I want to say the answer?
I don't think so, it spoils the fun, no?
it's not zero or 1/2 though

Answer 9

a better way would be

\[\huge e^{\ln n - \frac{1}{n}*ln n!}\]

Answer 10

yes it would :)

Answer 11

i am only gonna solve the power now
\[\ln n - \frac{1}{n} ( \ln n + \ln (n-1) + \ldots + \ln 0!) \]

Answer 12

interesting answer

Answer 13

can't you only do that e^blah trick for certain forms though? is this one of them?

Answer 14

you found the answer imran?

Answer 15

yes, but don't know how to get it

Answer 16

right, my problem exactly
Wolfram won't help me differentiate n! much, so I can't do l'Hospital...

Answer 17

product of n terms as n->infinity is candidate for the e^ln(stuff) trick

Answer 18

yes but I think it needs to be of the form\[1^{\infty}\]or something, right? I can't get it like that.

Answer 19

how about ( n^n / n! ) ^ (1/n) = e ^ ln( same ) = e ^ ( (summation ln(k)) / n ) if that makes sense to you.

Answer 20

which summation?

Answer 21

and like I said, I'm not sure of the rules on when you can do the e^(something) trick. not always, right?

Answer 22

turing the trick you are referring to is a special case.

\[e^{\log_e{x}} = x \]or \[n^{\log_n x} = x \]is a general case

Answer 23

I know that, I'm asking about Broken's proposed use of it here. Is it allowed yet, or do we need to alter the form first?

Answer 24

\[n / \sqrt[n]{n!} = \sqrt[n]{ n^n / n! } = e^{\ln{\sqrt[n]{ n^n / n! }}} = e^{(\ln({n^n/n!}))/n}\]
hasn't actually used any passing of limits through yet

Answer 25

Hence e to the power of \[\frac{1}{N}\sum_{k=1}^{N}\ln(\frac{N}{k})\]

Answer 26

I think it is easier to see as \[-\frac{1}{N}\sum_{k=1}^{N}\ln(\frac{k}{N})\]

Answer 27

i don't like factorials in limit's problem

Answer 28

compare that to the average value of the natural log between 0 and 1, or the integral of ln(x) from 0 to 1 ... Hence my assertion that the answer to the original limit was e^1 = e.

Answer 29

(Integral of ln(x) using evenly partitioned Riemann sums I mean)

Answer 30

\[+1 = -\int\limits_{0}^{1}\ln(x)dx = \lim_{n->\infty}-\frac{1}{n}\sum_{k=1}^{n}\ln(\frac{k}{n})\]
assuming that I typeset that correctly.

Answer 31

Nice, I'm convinced. Did you know the answer beforehand?

Answer 32

Broken you know when you wrote:
\[-\frac{1}{n}\sum_{k=1}^{n}\ln \left(\frac{n}{k}\right)\]

could you possibly explain why you got this instead of:
\[-\frac{1}{n}\sum_{k=1}^{n}\ln \left(\frac{n^n}{k}\right)\]
Assuming you derived it from: \[\frac{1}{n} \ln \left(\frac{n^n}{n!}\right)\]

Answer 33

sorry I didn't mean to write the minus there

Answer 34

\[\ln(\frac{n^n}{n!}) = \ln(\frac{n}{n})(\frac{n}{n-1})(\frac{n}{n-2})...(\frac{n}{1})=\ln(\frac{n}{n})+\ln(\frac{n}{n-1})+...+\ln(\frac{n}{1})=\sum_{k=1}^{n}\ln(\frac{n}{k})\]