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myko Group TitleBest ResponseYou've already chosen the best response.0
maybe like this: \[\sqrt[n]{n!}=\sqrt[n]{n(n1)(n2)\cdots1} =\sqrt[n]{n}\sqrt[n]{n1}\cdots \sqrt[n]{1}=1\] so limit is equal to 0
 2 years ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.0
Are you sure that \[\lim_{n\rightarrow\infty}\sqrt[n]{n!}=1\]
 2 years ago

myko Group TitleBest ResponseYou've already chosen the best response.0
look the steps from my comment befor. It looks ok
 2 years ago

myko Group TitleBest ResponseYou've already chosen the best response.0
all the roots at the right hand side: \[\sqrt[n]{n}=\sqrt[n]{n1}=\cdots=\sqrt[n]{1}=1\]
 2 years ago

myko Group TitleBest ResponseYou've already chosen the best response.0
so their product too
 2 years ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.0
No, it's not ok. Because you multiply an infinite quantity or 1. As we know \(1^{\infty}={}?\).
 2 years ago

myko Group TitleBest ResponseYou've already chosen the best response.0
\[1^{\infty} =1*1*\cdots*1=1\]
 2 years ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.0
Very nice. What can you say about this pretty limit? \[\lim_{n\rightarrow\infty}\left(1+\frac1n\right)^n\]It is \(1^{\infty}\).
 2 years ago

myko Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

myko Group TitleBest ResponseYou've already chosen the best response.0
in this case there is just number one multiplyed infinitly many times. And it happens after the limit was taken
 2 years ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.0
Ok. What about this? \[\lim_{n\rightarrow\infty}\sum_{k=1}^n\frac1n\]Is it 0?
 2 years ago

myko Group TitleBest ResponseYou've already chosen the best response.0
this is a harmonic series. It is not convergent
 2 years ago

myko Group TitleBest ResponseYou've already chosen the best response.0
your 1º question was about sequences
 2 years ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.0
Look at the denominator carefully please. I hopr you will try to get what I'm saying.
 2 years ago

myko Group TitleBest ResponseYou've already chosen the best response.0
sry, but i don't
 2 years ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.0
Can you find this? \[\lim_{n\rightarrow\infty}\sum_{k=1}^n\frac1n\]
 2 years ago

myko Group TitleBest ResponseYou've already chosen the best response.0
another way to try this: \[\lim \sqrt[n]{\frac{n!}{n^n}} = 0\]
 2 years ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.0
Can you show the way you solve it?
 2 years ago

myko Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.0
@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
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
@mahmit2012 a little help here?
 2 years ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.3
dw:1351370532404:dw
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
very nice, my turn...
 2 years ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.3
dw:1351370740820:dw
 2 years ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.3
dw:1351370823511:dw
 2 years ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.3
dw:1351370856590:dw
 2 years ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
\[{\sqrt[n]{n!}\over n}=\exp\left(\frac1n\ln(n!)\ln n\right)=\exp\left({n\ln nn+O(n)n\ln n\over n}\right)\]\[=\exp(1)=\frac1e\]
 2 years ago

myko Group TitleBest ResponseYou've already chosen the best response.0
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 \[\lim \sqrt[n]{a_{n}} = \frac{1}{e}\]
 2 years ago

myko Group TitleBest ResponseYou've already chosen the best response.0
@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
 2 years ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.0
Nice. But I have one more interesting method to find it. \[\lim_{n\rightarrow\infty} \frac{\sqrt[n]{n!}}{n}=\lim_{n\rightarrow\infty}\sqrt[n]{\frac{n!}{n^n}}=\lim_{n\rightarrow\infty}\sqrt[n]{\frac1n\cdot\frac2n\cdots\frac n n}=A\]\[\ln A=\lim_{n\rightarrow\infty}\frac1n(\ln\frac1n+\ln\frac2n+\ldots+\ln\frac n n)=\int_0^1\ln x dx=1\]\[\lim_{n\rightarrow\infty} \frac{\sqrt[n]{n!}}{n}=A=e^{1}=\frac1e\]
 2 years ago
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