## klimenkov 3 years ago $\lim_{n\rightarrow\infty}\frac{\sqrt[n]{n!}}{n}$

1. myko

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

2. myko

@klimenkov

3. klimenkov

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

4. myko

look the steps from my comment befor. It looks ok

5. myko

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

6. myko

so their product too

7. klimenkov

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

8. myko

$1^{\infty} =1*1*\cdots*1=1$

9. klimenkov

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

10. myko

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

11. myko

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

12. klimenkov

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

13. myko

this is a harmonic series. It is not convergent

14. myko

15. klimenkov

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

16. myko

sry, but i don't

17. klimenkov

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

18. myko

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

19. myko

infinity

20. klimenkov

Can you show the way you solve it?

21. myko

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

22. klimenkov

@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

23. TuringTest

@mahmit2012 a little help here?

24. mahmit2012

|dw:1351370532404:dw|

25. TuringTest

very nice, my turn...

26. mahmit2012

|dw:1351370740820:dw|

27. mahmit2012

|dw:1351370823511:dw|

28. mahmit2012

|dw:1351370856590:dw|

29. TuringTest

${\sqrt[n]{n!}\over n}=\exp\left(\frac1n\ln(n!)-\ln n\right)=\exp\left({n\ln n-n+O(n)-n\ln n\over n}\right)$$=\exp(-1)=\frac1e$

30. mukushla

*

31. myko

ya I was wrong. Here is another way to solve it. As we know root test is stronger than cuotient test, so the folowing inequality holds: $\lim \inf \frac{a_{n+1}}{a_{n}}\leq\lim \inf \sqrt[n]{a_{n}}\leq \lim \sup \sqrt[n]{a_{n}} \leq \lim \sup\frac{a_{n+1}}{a_{n}}$ let $a_{n} = \frac{n!}{n^{n}}$ then $\frac{a_{n+1}}{a_{n}}=\frac{1}{(1+\frac{1}{n})^{n}}=\frac{1}{e}$ this means that $\lim \sqrt[n]{a_{n}} = \frac{1}{e}$

32. myko

@mahmit2012 $\lim \frac{a_{n+1}}{a_{n}}=\lim \sqrt[n]{a_{n}}$ only if a_n is convergent, what is not implied in this question

33. klimenkov

Nice. But I have one more interesting method to find it. $\lim_{n\rightarrow\infty} \frac{\sqrt[n]{n!}}{n}=\lim_{n\rightarrow\infty}\sqrt[n]{\frac{n!}{n^n}}=\lim_{n\rightarrow\infty}\sqrt[n]{\frac1n\cdot\frac2n\cdots\frac n n}=A$$\ln A=\lim_{n\rightarrow\infty}\frac1n(\ln\frac1n+\ln\frac2n+\ldots+\ln\frac n n)=\int_0^1\ln x dx=-1$$\lim_{n\rightarrow\infty} \frac{\sqrt[n]{n!}}{n}=A=e^{-1}=\frac1e$