klimenkov
  • klimenkov
\[\lim_{n\rightarrow\infty}\frac{\sqrt[n]{n!}}{n}\]
Calculus1
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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
anonymous
  • anonymous
@klimenkov
klimenkov
  • klimenkov
Are you sure that \[\lim_{n\rightarrow\infty}\sqrt[n]{n!}=1\]

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anonymous
  • anonymous
look the steps from my comment befor. It looks ok
anonymous
  • anonymous
all the roots at the right hand side: \[\sqrt[n]{n}=\sqrt[n]{n-1}=\cdots=\sqrt[n]{1}=1\]
anonymous
  • anonymous
so their product too
klimenkov
  • klimenkov
No, it's not ok. Because you multiply an infinite quantity or 1. As we know \(1^{\infty}={}?\).
anonymous
  • anonymous
\[1^{\infty} =1*1*\cdots*1=1\]
klimenkov
  • klimenkov
Very nice. What can you say about this pretty limit? \[\lim_{n\rightarrow\infty}\left(1+\frac1n\right)^n\]It is \(1^{\infty}\).
anonymous
  • anonymous
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
anonymous
  • anonymous
in this case there is just number one multiplyed infinitly many times. And it happens after the limit was taken
klimenkov
  • klimenkov
Ok. What about this? \[\lim_{n\rightarrow\infty}\sum_{k=1}^n\frac1n\]Is it 0?
anonymous
  • anonymous
this is a harmonic series. It is not convergent
anonymous
  • anonymous
your 1ยบ question was about sequences
klimenkov
  • klimenkov
Look at the denominator carefully please. I hopr you will try to get what I'm saying.
anonymous
  • anonymous
sry, but i don't
klimenkov
  • klimenkov
Can you find this? \[\lim_{n\rightarrow\infty}\sum_{k=1}^n\frac1n\]
anonymous
  • anonymous
another way to try this: \[\lim \sqrt[n]{\frac{n!}{n^n}} = 0\]
anonymous
  • anonymous
infinity
klimenkov
  • klimenkov
Can you show the way you solve it?
anonymous
  • anonymous
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
klimenkov
  • 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
TuringTest
  • TuringTest
@mahmit2012 a little help here?
anonymous
  • anonymous
|dw:1351370532404:dw|
TuringTest
  • TuringTest
very nice, my turn...
anonymous
  • anonymous
|dw:1351370740820:dw|
anonymous
  • anonymous
|dw:1351370823511:dw|
anonymous
  • anonymous
|dw:1351370856590:dw|
TuringTest
  • TuringTest
\[{\sqrt[n]{n!}\over n}=\exp\left(\frac1n\ln(n!)-\ln n\right)=\exp\left({n\ln n-n+O(n)-n\ln n\over n}\right)\]\[=\exp(-1)=\frac1e\]
anonymous
  • anonymous
*
anonymous
  • anonymous
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 \[\lim \sqrt[n]{a_{n}} = \frac{1}{e}\]
anonymous
  • anonymous
@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
klimenkov
  • klimenkov
Nice. But I have one more interesting method to find it. \[\lim_{n\rightarrow\infty} \frac{\sqrt[n]{n!}}{n}=\lim_{n\rightarrow\infty}\sqrt[n]{\frac{n!}{n^n}}=\lim_{n\rightarrow\infty}\sqrt[n]{\frac1n\cdot\frac2n\cdots\frac n n}=A\]\[\ln A=\lim_{n\rightarrow\infty}\frac1n(\ln\frac1n+\ln\frac2n+\ldots+\ln\frac n n)=\int_0^1\ln x dx=-1\]\[\lim_{n\rightarrow\infty} \frac{\sqrt[n]{n!}}{n}=A=e^{-1}=\frac1e\]

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