Question

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

Answer 1

maybe like this:
\[\sqrt[n]{n!}=\sqrt[n]{n(n-1)(n-2)\cdots1} =\sqrt[n]{n}\sqrt[n]{n-1}\cdots \sqrt[n]{1}=1\]
so limit is equal to 0

Answer 2

@klimenkov

Answer 3

Are you sure that
\[\lim_{n\rightarrow\infty}\sqrt[n]{n!}=1\]

Answer 4

look the steps from my comment befor. It looks ok

Answer 5

all the roots at the right hand side:
\[\sqrt[n]{n}=\sqrt[n]{n-1}=\cdots=\sqrt[n]{1}=1\]

Answer 6

so their product too

Answer 7

No, it's not ok. Because you multiply an infinite quantity or 1. As we know \(1^{\infty}={}?\).

Answer 8

\[1^{\infty} =1*1*\cdots*1=1\]

Answer 9

Very nice. What can you say about this pretty limit? \[\lim_{n\rightarrow\infty}\left(1+\frac1n\right)^n\]It is \(1^{\infty}\).

Answer 10

this happens when you talk about functions. The reason of indetermination of 1^infinity is because of that. But in this case there are no functions involved. That's my point

Answer 11

in this case there is just number one multiplyed infinitly many times. And it happens after the limit was taken

Answer 12

Ok. What about this?
\[\lim_{n\rightarrow\infty}\sum_{k=1}^n\frac1n\]Is it 0?

Answer 13

this is a harmonic series. It is not convergent

Answer 14

your 1º question was about sequences

Answer 15

Look at the denominator carefully please. I hopr you will try to get what I'm saying.

Answer 16

sry, but i don't

Answer 17

Can you find this?
\[\lim_{n\rightarrow\infty}\sum_{k=1}^n\frac1n\]

Answer 18

another way to try this:
\[\lim \sqrt[n]{\frac{n!}{n^n}} = 0\]

Answer 19

infinity

Answer 20

Can you show the way you solve it?

Answer 21

I don't remmeber the formal proof of n!/n^n =0, but it's evident, if you try a few first terms of this sequence. There are some posts about it if you google a bit

Answer 22

@myko, see this and tell me what is your mistake?
http://www.wolframalpha.com/input/?i=Limit+%28n!%29^%281%2Fn%29%2Fn+n-%3Einfinity

Answer 23

@mahmit2012 a little help here?