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mykoBest 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
 one year ago

klimenkovBest ResponseYou've already chosen the best response.0
Are you sure that \[\lim_{n\rightarrow\infty}\sqrt[n]{n!}=1\]
 one year ago

mykoBest ResponseYou've already chosen the best response.0
look the steps from my comment befor. It looks ok
 one year ago

mykoBest ResponseYou've already chosen the best response.0
all the roots at the right hand side: \[\sqrt[n]{n}=\sqrt[n]{n1}=\cdots=\sqrt[n]{1}=1\]
 one year ago

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

mykoBest ResponseYou've already chosen the best response.0
\[1^{\infty} =1*1*\cdots*1=1\]
 one year ago

klimenkovBest 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 \(1^{\infty}\).
 one year ago

mykoBest 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
 one year ago

mykoBest 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
 one year ago

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

mykoBest ResponseYou've already chosen the best response.0
this is a harmonic series. It is not convergent
 one year ago

mykoBest ResponseYou've already chosen the best response.0
your 1º question was about sequences
 one year ago

klimenkovBest 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.
 one year ago

klimenkovBest ResponseYou've already chosen the best response.0
Can you find this? \[\lim_{n\rightarrow\infty}\sum_{k=1}^n\frac1n\]
 one year ago

mykoBest ResponseYou've already chosen the best response.0
another way to try this: \[\lim \sqrt[n]{\frac{n!}{n^n}} = 0\]
 one year ago

klimenkovBest ResponseYou've already chosen the best response.0
Can you show the way you solve it?
 one year ago

mykoBest 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
 one year ago

klimenkovBest 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
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
@mahmit2012 a little help here?
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.3
dw:1351370532404:dw
 one year ago

TuringTestBest ResponseYou've already chosen the best response.0
very nice, my turn...
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.3
dw:1351370740820:dw
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.3
dw:1351370823511:dw
 one year ago

mahmit2012Best ResponseYou've already chosen the best response.3
dw:1351370856590:dw
 one year ago

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

mykoBest 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 \[\lim \sqrt[n]{a_{n}} = \frac{1}{e}\]
 one year ago

mykoBest 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
 one year ago

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