Here's the question you clicked on:
klimenkov
\[\lim_{n\rightarrow\infty}\frac{\sqrt[n]{n!}}{n}\]
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
Are you sure that \[\lim_{n\rightarrow\infty}\sqrt[n]{n!}=1\]
look the steps from my comment befor. It looks ok
all the roots at the right hand side: \[\sqrt[n]{n}=\sqrt[n]{n-1}=\cdots=\sqrt[n]{1}=1\]
No, it's not ok. Because you multiply an infinite quantity or 1. As we know \(1^{\infty}={}?\).
\[1^{\infty} =1*1*\cdots*1=1\]
Very nice. What can you say about this pretty limit? \[\lim_{n\rightarrow\infty}\left(1+\frac1n\right)^n\]It is \(1^{\infty}\).
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
in this case there is just number one multiplyed infinitly many times. And it happens after the limit was taken
Ok. What about this? \[\lim_{n\rightarrow\infty}\sum_{k=1}^n\frac1n\]Is it 0?
this is a harmonic series. It is not convergent
your 1º question was about sequences
Look at the denominator carefully please. I hopr you will try to get what I'm saying.
Can you find this? \[\lim_{n\rightarrow\infty}\sum_{k=1}^n\frac1n\]
another way to try this: \[\lim \sqrt[n]{\frac{n!}{n^n}} = 0\]
Can you show the way you solve it?
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
@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
@mahmit2012 a little help here?
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very nice, my turn...
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\[{\sqrt[n]{n!}\over n}=\exp\left(\frac1n\ln(n!)-\ln n\right)=\exp\left({n\ln n-n+O(n)-n\ln n\over n}\right)\]\[=\exp(-1)=\frac1e\]
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 \[\lim \sqrt[n]{a_{n}} = \frac{1}{e}\]
@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
Nice. But I have one more interesting method to find it. \[\lim_{n\rightarrow\infty} \frac{\sqrt[n]{n!}}{n}=\lim_{n\rightarrow\infty}\sqrt[n]{\frac{n!}{n^n}}=\lim_{n\rightarrow\infty}\sqrt[n]{\frac1n\cdot\frac2n\cdots\frac n n}=A\]\[\ln A=\lim_{n\rightarrow\infty}\frac1n(\ln\frac1n+\ln\frac2n+\ldots+\ln\frac n n)=\int_0^1\ln x dx=-1\]\[\lim_{n\rightarrow\infty} \frac{\sqrt[n]{n!}}{n}=A=e^{-1}=\frac1e\]