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Serie convergence

  1. Jun 8, 2012 #1
    1. The problem statement, all variables and given/known data

    Hi. I'm trying to solve a serie:

    Ʃ12n+1*(n+1)! / (n+1)n+1

    2. Relevant equations



    3. The attempt at a solution

    I tried solving it with Cauchy's method, but it failed. I also tried using d'alembert criterion, which game me the answer 2, so it should be divergent.
    However in the solutions it says it's convergent. If anyone could check my solution and see if I did anything wrong or if it's a simple error in the solutions, I'd appreciate!
    21e5g5g.png
     
  2. jcsd
  3. Jun 8, 2012 #2
    You've made a small mistake in the last step. When you have,

    [tex]\lim_{n\to \infty} 2 \left ( \frac{1+1/n}{1+2/n} \right ) ^{n+1}[/tex]

    You cant directly write it to be equal to 2,because the term in the bracket is of the indeterminate form [itex]1^{\infty}[/itex]
     
  4. Jun 8, 2012 #3
    Oh, I see... I tought I could say it was equal to 1... Any ideas then on how to solve it?
     
  5. Jun 8, 2012 #4
    Um, that is a standard limit of the form,

    [tex]\lim_{x\to \infty} (1+\frac{1}{x})^x[/tex]

    or, more generally,

    [tex]\lim (1+f(x))^{g(x)}[/tex]
    As f(x) tends to zero and g(x) tends to infinity.

    Did you not learn about these?
     
  6. Jun 8, 2012 #5
    I don't think so. I studied
    [tex]\lim_{x\to \infty} (1+\frac{1}{x})^x[/tex]
    and I know it tends to e, but I don't know how it helps me in this case.I can't rewrite
    [tex]\lim_{n\to \infty} 2 \left ( \frac{1+1/n}{1+2/n} \right ) ^{n+1}[/tex]
    as
    [tex]\lim_{x\to \infty} (1+\frac{1}{x})^x[/tex]
     
  7. Jun 8, 2012 #6
    You should ask your teacher to explain these types of limits, then.

    To find

    [tex]\lim_{x\to \infty}(1+f(x))^{g(x)}[/tex]

    Such that f(x) tends to 0 and g(x) tends to infinity is,

    [tex]e^{\lim{x\to \infty}f(x) \cdot g(x)}[/tex]

    The limit of [itex](1+1/x)^x[/itex] as [itex]x\to \infty[/itex] is a particular case(sometimes, even used as the definition of e) of this general form.
     
  8. Jun 8, 2012 #7
    Thank you very much :)
     
  9. Jun 8, 2012 #8
    Hi Jalo, (Hi Infinitum)
    First, you could have made your life easier by starting with a simpler index
    (you have n+2 terms over n+1 terms that you could instead put as n+1 terms over n terms)
    Then that would lead you to think about the limit of 2(1+1/n)^n
    I'm saying think about instead of find out, because as Infinitum told you, this limit can be tricky, and you learnt about it or not.
    But what is important is the original question, you were not asked to evaluate precisely this limit, you just wanted to check D'Alembert's rule.
    Now (1+1/n)^n is hard if you are not properly prepared to deal with it,
    but pretend you are not, do you really need to be ?
    Whatever this limit is, who knows, it could go to infinity, one thing is certain, it is >=1, that is all you need, since you multiply it by 2, to know that your rule for divergence is satisfied.

    Cheers...
     
  10. Jun 8, 2012 #9
    Hello Infinitum,

    I have never heard of this. Does it has a particular name? I'm trying to look on the internet, but I don't find anything related to that. Do you have a link that you could share?
     
  11. Jun 8, 2012 #10

    Ray Vickson

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    Science Advisor
    Homework Helper

    Use Stirling's Formula: [itex] N! \sim \sqrt{2 \pi} N^{N}\sqrt{N} e^{-N}[/itex] for large N. Here, the symbol "~" means that the ratio of the two sides approaches 1 as N approaches infinity.

    RGV
     
  12. Jun 8, 2012 #11
    It is the limit of e only a bit modified.

    [tex]\lim_{x\to a} (1+f(x))^{g(x)}[/tex]

    Can be re-written as

    [tex]\lim_{x\to a} \left (1+f(x))^{\frac{1}{f(x)}} \right)^{g(x)f(x)} [/tex]


    I think you should be able to understand it from here
     
  13. Jun 8, 2012 #12
    I see! Really interesting - I actually didn't saw the exponent and the (1/x) inside the parenthesis (for the particular case, e) as to independent functions...
     
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