# I Upper Bound of a Summation

Tags:
1. Oct 18, 2016

### 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.

2. Oct 19, 2016

### 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.

3. Oct 19, 2016

### mooncrater

Thanks ! :)

4. Oct 19, 2016

### 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.

5. Oct 19, 2016

### Svein

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

6. Oct 19, 2016

### Staff: Mentor

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

7. Oct 19, 2016

### mooncrater

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