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Math Help - Convergence value of sum_n=0:Inf ( x^n / n! )^k

  1. #1
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    Convergence value of sum_n=0:Inf ( x^n / n! )^k

    Is there any way to compute analytically the convergence value of the series:



    We know that for k=1, the convergence value is .
    For k->+Inf the series converges to 2.
    For k>1 it is possible to verify that the series converges. The problem is to determine analytically this value.
    My questions are:
    1. Is there any way to do this?
    2. Is there any relation with the Napier number?


    Thank in advance!
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  2. #2
    mfb
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    Re: Convergence value of sum_n=0:Inf ( x^n / n! )^k

    For k->+Inf the series converges to 2.
    For lambda=3, the sum is (\frac{1}{1})^k + (\frac{3}{2})^k + (\frac{9}{6})^k + (\frac{27}{24})^k + \sum_i a_i^k where all ai<1. With k -> inf, this sum vanishes, but the first terms and their sum diverge.

    In the limit k->inf:
    With lambda=-2, it diverges (oscillates between 0 for odd k and 2 for even k)
    With -2<lambda<2, it converges to 1.
    With lambda=2, it converges to 2.
    With lambda>2, it diverges

    With lambda< -2, I would expect divergence, too.

    The series itself converges for every k>1, I agree. For every lambda, only a finite number of terms is larger than 1 and with k=1 it converges, therefore taking the k'th power does not change convergence. However, the limit k->inf depends on lambda.

    The problem is to determine analytically this value.
    No useful idea, sorry.
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    Re: Convergence value of sum_n=0:Inf ( x^n / n! )^k

    Quote Originally Posted by mfb View Post
    For lambda=3, the sum is (\frac{1}{1})^k + (\frac{3}{2})^k + (\frac{9}{6})^k + (\frac{27}{24})^k + \sum_i a_i^k where all ai<1. With k -> inf, this sum vanishes, but the first terms and their sum diverge.

    In the limit k->inf:
    With lambda=-2, it diverges (oscillates between 0 for odd k and 2 for even k)
    With -2<lambda<2, it converges to 1.
    With lambda=2, it converges to 2.
    With lambda>2, it diverges

    With lambda< -2, I would expect divergence, too.

    The series itself converges for every k>1, I agree. For every lambda, only a finite number of terms is larger than 1 and with k=1 it converges, therefore taking the k'th power does not change convergence. However, the limit k->inf depends on lambda.


    No useful idea, sorry.
    Thnx for your reply. There was a lapsus in my first message. When I said that the series converges to 2 for k->Inf I ment for lambda = 1, so we have the following limit:

    \lim_{k\to+\infty}\sum_{n=0}^{+\infty}\left(\frac{  1}{n!}\right)^k

    by putting in evidence the first two terms we have:

    \lim_{k\to+\infty}\left[ (1/0!)^k + (1/1!)^k+\sum_{n=2}^{+\infty}\left(\frac{1}{n!}\right  )^k\right ]

    equivalently:

    (1/0!)^k + (1/1!)^k+\sum_{n=2}^{+\infty}\lim_{k\to+\infty} \left(\frac{1}{n!}\right)^k

    so:

    \lim_{k\to+\infty}\sum_{n=0}^{+\infty}\left(\frac{  1}{n!}\right)^k=2

    but this contradicts your tesis that for \lambda\in]-2;+2[ the series converges to 1.

    I will be satisfied even if it is possible to find an analytical solution for the particular case lambda=1.
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  4. #4
    mfb
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    Re: Convergence value of sum_n=0:Inf ( x^n / n! )^k

    Oh sorry, I messed up the terms.

    New try:
    1/1 + lambda^k/1 + lambda^(2k)/2^k + ...

    This diverges for lambda>1. Lambda=1 gives a limit of 2, |lambda|<1 gives a limit of 1.
    lambda = -1 gives divergence, lambda < -1 would need more analysis.
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