What is the Upper Bound of this Summation?

[mooncrater]

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?
Moon.

 

[Fightfish]

If you rewrite the summation in terms of $\lambda \equiv (n-i)$, you get a partial sum of the exponential series:
\left(\frac{1}{m^{n-1}}\right)\,\sum_{\lambda = 0}^{n-k} \frac{m^{\lambda}}{\lambda!},
which well can be bounded by the exponential or written in terms of the incomplete gamma function.

 

[mooncrater]

Thanks ! :)

 

[mooncrater]

Sorry, but I'm facing a problem again. I found about incomplete gamma functions on wikipedia:
https://en.wikipedia.org/wiki/Incomplete_gamma_function

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 :
$$ \gamma (s,z) = z^{s} \Gamma (s) \gamma *(s,z) $$
Thus,
$$\gamma *(s,z) =\frac { \gamma(s,z)} {z^{s} \Gamma(s) } $$
since $s=0$, therefore :
$$\gamma *(0,z) = \frac{\gamma(0,z)} {1* \Gamma(0)} $$
But, according to this link :
https://en.wikipedia.org/wiki/Particular_values_of_the_Gamma_function
Gamma function is not defined for non-positive integers.
So, how do I tackle this $\Gamma(0)$ ?
Moon.

 

[Svein]

Gamma function is not defined for non-positive integers.
So, how do I tackle this Γ(0)\Gamma(0) ?

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

 

[mfb]

Huh?
$$\gamma *(0,z) = e^{-z} \sum_{k=0}^{\infty} \frac {z^{k}}{ k!} = e^{-z} e^z = 1$$
Done?

 

[mooncrater]

Huh?
$$\gamma *(0,z) = e^{-z} \sum_{k=0}^{\infty} \frac {z^{k}}{ k!} = e^{-z} e^z = 1$$
Done?

Sorry, forgot about the expansion of the majestic $e^{x}$ itself.
Moon.