Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

klimenkov

\[\lim_{n\rightarrow\infty}\frac{\sqrt[n]{n!}}{n}\]

  • one year ago
  • one year ago

  • This Question is Closed
  1. myko
    Best Response
    You've already chosen the best response.
    Medals 0

    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

    • one year ago
  2. myko
    Best Response
    You've already chosen the best response.
    Medals 0

    @klimenkov

    • one year ago
  3. klimenkov
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  4. myko
    Best Response
    You've already chosen the best response.
    Medals 0

    look the steps from my comment befor. It looks ok

    • one year ago
  5. myko
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  6. myko
    Best Response
    You've already chosen the best response.
    Medals 0

    so their product too

    • one year ago
  7. klimenkov
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  8. myko
    Best Response
    You've already chosen the best response.
    Medals 0

    \[1^{\infty} =1*1*\cdots*1=1\]

    • one year ago
  9. klimenkov
    Best Response
    You've already chosen the best response.
    Medals 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
  10. myko
    Best Response
    You've already chosen the best response.
    Medals 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
  11. myko
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  12. klimenkov
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  13. myko
    Best Response
    You've already chosen the best response.
    Medals 0

    this is a harmonic series. It is not convergent

    • one year ago
  14. myko
    Best Response
    You've already chosen the best response.
    Medals 0

    your 1º question was about sequences

    • one year ago
  15. klimenkov
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  16. myko
    Best Response
    You've already chosen the best response.
    Medals 0

    sry, but i don't

    • one year ago
  17. klimenkov
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  18. myko
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  19. myko
    Best Response
    You've already chosen the best response.
    Medals 0

    infinity

    • one year ago
  20. klimenkov
    Best Response
    You've already chosen the best response.
    Medals 0

    Can you show the way you solve it?

    • one year ago
  21. myko
    Best Response
    You've already chosen the best response.
    Medals 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
  22. klimenkov
    Best Response
    You've already chosen the best response.
    Medals 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
  23. TuringTest
    Best Response
    You've already chosen the best response.
    Medals 0

    @mahmit2012 a little help here?

    • one year ago
  24. mahmit2012
    Best Response
    You've already chosen the best response.
    Medals 3

    |dw:1351370532404:dw|

    • one year ago
  25. TuringTest
    Best Response
    You've already chosen the best response.
    Medals 0

    very nice, my turn...

    • one year ago
  26. mahmit2012
    Best Response
    You've already chosen the best response.
    Medals 3

    |dw:1351370740820:dw|

    • one year ago
  27. mahmit2012
    Best Response
    You've already chosen the best response.
    Medals 3

    |dw:1351370823511:dw|

    • one year ago
  28. mahmit2012
    Best Response
    You've already chosen the best response.
    Medals 3

    |dw:1351370856590:dw|

    • one year ago
  29. TuringTest
    Best Response
    You've already chosen the best response.
    Medals 0

    \[{\sqrt[n]{n!}\over n}=\exp\left(\frac1n\ln(n!)-\ln n\right)=\exp\left({n\ln n-n+O(n)-n\ln n\over n}\right)\]\[=\exp(-1)=\frac1e\]

    • one year ago
  30. mukushla
    Best Response
    You've already chosen the best response.
    Medals 0

    *

    • one year ago
  31. myko
    Best Response
    You've already chosen the best response.
    Medals 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
  32. myko
    Best Response
    You've already chosen the best response.
    Medals 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
  33. klimenkov
    Best Response
    You've already chosen the best response.
    Medals 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
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.