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Series convergence for certain values of p

  1. May 25, 2014 #1
    1. The problem statement, all variables and given/known data

    For which integer values of p does the following series converge:

    [itex]\sum_{n=|p|}^{∞}{2^{pn} (n+p)! \over(n+p)^n}[/itex]


    2. Relevant equations



    3. The attempt at a solution

    I'm trying to apply the generalised ratio test but get down to this stage where I'm not sure what to do next.

    I have [itex]\lim{2^p(n+p)^n \over(n+1+p)}[/itex] so far from attempting to do the generalised ratio test.
     
  2. jcsd
  3. May 25, 2014 #2

    BvU

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    Hello Yambo, and welcome to PF.

    Could you check your denominator ?

    [edit] And: if p≤0, what is the first term in the summation ?
     
    Last edited: May 25, 2014
  4. May 25, 2014 #3
    Hi BvU,

    I've checked it. Got the sheet in front of me and it is definitely that.

    (unless you mean I have something wrong in my attempt. )
     
  5. May 25, 2014 #4

    BvU

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    Well, could you show the steps that lead to that ratio?
    I do get something else in the denominator.
     
  6. May 25, 2014 #5
    I shall omit the modulus signs and the lim in the equations.
    [itex]
    2^{p(n+1)}((n+1)+p)!\over((n+1)+p)^{(n+1)}
    [/itex]
    multiplied by
    [itex]
    (n+p)^n\over 2^{pn}(n+p)!
    [/itex]

    [itex]
    2^{pn}2^{p}(n+1+p)(n+p)!(n+p)^n \over (n+1+p)^n(n+1+p)2^{pn}(n+p)!
    [/itex]

    then cancelling gives me

    [itex]

    2^{p}(n+p)^n \over (n+1+p)^n
    [/itex]

    Is there something wrong with my cancellation?

    edit* Oh I realised I forgot a ^n in my denominator. oops. I did have it down on paper though. not sure how to go from this stage.
     
    Last edited: May 25, 2014
  7. May 26, 2014 #6

    BvU

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    Good thing you found that. Big difference.

    Now we have to take the limit for n to ##\infty## and see what comes out. Not trivial, but not impossible either. 'Not sure' and 'don't know what to do' aren't good enough according to PF rules. Just giving you the next step would nullify the learning experience :smile:. So: What do you have in your toolbox to deal with this kind of limits ?

    Coming back to the modulus sign: see post #2.
     
  8. May 26, 2014 #7
    Well considering I've been doing this problem for 2 days I think I'm missing something from my tool box. Could you give me a clue? The furthest I've gotten is to factor the 2^p out and then simply put everything in brackets to the power of n. I'm not even sure if this is the right way to go because it leads to a dead end still, unless you can use standard limit 2 and say everything in the brackets is equal to r... but I have a very strong feeling you can't do that.

    As for the first term, if n=0, we get p!
     
    Last edited: May 26, 2014
  9. May 26, 2014 #8

    ehild

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    Putting everything in brackets to the power of n is a good idea... You have a fraction inside the brackets.

    Use the identity [tex]\frac{a+b}{a+b+1}=\frac{1}{1+\frac{1}{a+b}}[/tex]

    ehild
     
  10. May 26, 2014 #9

    pasmith

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    I would recommend using Stirling's approximation: [tex]
    (n + p)! \sim \sqrt{2\pi(n + p)}\left(\frac{n + p}{e}\right)^{n+p}[/tex]

    The first term is not [itex]n = 0[/itex]; the first term is [itex]n = |p|[/itex]. If [itex]p \leq 0[/itex] and [itex]n = |p|[/itex], then [itex]n = -p[/itex] and [itex](n + p) = (-p + p) = 0[/itex]. This is a problem.
     
  11. May 26, 2014 #10

    BvU

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    PA helps you with the modulus thing. I don't know why one should want to bring back in the factorials, but I haven't checked that out.

    I figured you might end up studying something in the direction of ##\lim_{n\rightarrow\infty}\, \left(1+{x \over n}\right )^n ##. Now there's a hint !

    Kudos for your persistence!
     
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