## klimenkov Group Title $\lim_{n\rightarrow\infty}\frac{\sqrt[n]{n!}}{n}$ one year ago one year ago

1. myko Group Title

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 Group Title

@klimenkov

3. klimenkov Group Title

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

4. myko Group Title

look the steps from my comment befor. It looks ok

5. myko Group Title

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

6. myko Group Title

so their product too

7. klimenkov Group Title

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

8. myko Group Title

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

9. klimenkov Group Title

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 Group Title

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 Group Title

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

12. klimenkov Group Title

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

13. myko Group Title

this is a harmonic series. It is not convergent

14. myko Group Title

15. klimenkov Group Title

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

16. myko Group Title

sry, but i don't

17. klimenkov Group Title

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

18. myko Group Title

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

19. myko Group Title

infinity

20. klimenkov Group Title

Can you show the way you solve it?

21. myko Group Title

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 Group Title

@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 Group Title

@mahmit2012 a little help here?

24. mahmit2012 Group Title

|dw:1351370532404:dw|

25. TuringTest Group Title

very nice, my turn...

26. mahmit2012 Group Title

|dw:1351370740820:dw|

27. mahmit2012 Group Title

|dw:1351370823511:dw|

28. mahmit2012 Group Title

|dw:1351370856590:dw|

29. TuringTest Group Title

${\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 Group Title

*

31. myko Group Title

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 Group Title

@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 Group Title

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$