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I Upper Bound of a Summation

  1. Oct 18, 2016 #1
    There is this summation, that I've been trying to solve, but am not able to do so. It is :
    $$\sum\limits_{i=k}^{n} \frac {1}{(n-i)! m^{i-1}}$$
    I would be happy to find it's upper bound too. So what I did was intensely naive. I made the denominator the minimum by making ##(n-i)! = 1## and ##m^{i-1} = m^{k-1}## (As that would give an upper bound too, but rather a loose one). That, as expected, didn't suffice my needs. Any ideas about how to solve this one?
  2. jcsd
  3. Oct 19, 2016 #2
    If you rewrite the summation in terms of ##\lambda \equiv (n-i)##, you get a partial sum of the exponential series:
    [tex]\left(\frac{1}{m^{n-1}}\right)\,\sum_{\lambda = 0}^{n-k} \frac{m^{\lambda}}{\lambda!}[/tex],
    which well can be bounded by the exponential or written in terms of the incomplete gamma function.
  4. Oct 19, 2016 #3
    Thanks ! :)
  5. Oct 19, 2016 #4
    Sorry, but I'm facing a problem again. I found about incomplete gamma functions on wikipedia:

    And I found that the upper incomplete gamma function, and the limiting function( defined as ##\gamma *##) suffice my needs. As it's given that :
    $$\gamma * (s,z) = e^{-z} \sum_{k=0}^{\infty} \frac {z^{k}}{\Gamma (s+k+1)} $$
    which is equivalent to:
    $$\gamma *(s,z) = e^{-z} \sum_{k=0}^{\infty} \frac {z^{k}}{(s+k)!} $$
    When we put ##s=0## then we get :
    $$\gamma *(0,z) = e^{-z} \sum_{k=0}^{\infty} \frac {z^{k}}{ k!} $$
    thus in order to find this summation, we need to find :
    $$\ {\gamma *(0,z)} {e^{z}} $$
    So, basically we've to find ##\gamma *(0,z)## .
    There is this Hargot's Theorem given in the wikipedia's page which says :
    $$ \gamma (s,z) = z^{s} \Gamma (s) \gamma *(s,z) $$
    $$\gamma *(s,z) =\frac { \gamma(s,z)} {z^{s} \Gamma(s) } $$
    since ##s=0##, therefore :
    $$\gamma *(0,z) = \frac{\gamma(0,z)} {1* \Gamma(0)} $$
    But, according to this link :
    Gamma function is not defined for non-positive integers.
    So, how do I tackle this ##\Gamma(0)## ?
  6. Oct 19, 2016 #5


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    Science Advisor

    Quoate from Ahlfors: Γ(z) is a meromorphic function with poles at z = 0, -1, -2, ... but without zeros.
  7. Oct 19, 2016 #6


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    Staff: Mentor

    $$\gamma *(0,z) = e^{-z} \sum_{k=0}^{\infty} \frac {z^{k}}{ k!} = e^{-z} e^z = 1$$
  8. Oct 19, 2016 #7
    Sorry, forgot about the expansion of the majestic ##e^{x}## itself.
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