Simplifying sums that contain factorials and have an index that doesn't start at 0. I'm trying to simplify the following sum:
$P(a)=\sum_{n=0}^{\infty}\dfrac{x^{n-a}}{(n-a)!}$
I'm tempted to turn this into some sort of exponential function because it closely ressembles the Maclaurin series $e^x=\sum_{n=0}^{\infty}\dfrac{x^n}{n!}$. Below is my attempt at this:
let $N=n-a$ so index now starts at $N=a$
$P(a)=\sum_{N=a}^{\infty}\dfrac{x^N}{N!}$
So now I have succeeded in making a function that is of the same form. Unfortunately, I don't really know what to do about the sum starting at N=a. Rewriting, I get:
$P(a)=\sum_{N=0}^{\infty}\dfrac{x^N}{N!}-\sum_{N=0}^a\dfrac{x^N}{N!}$
Now I definitely can write the first term as an exponential yaaay! But I'm still left with a finite sum that I have to subtract. So I guess what I'm really asking is what do I do about the second term?
Edit: It seems my first equation is incorrect. Here is the context surrounding my question:

Note: The variables were relabeled in my original question and constants factors were pulled out to make things a little easier on the eye.
I'm currently on part b). Subbing the given $P(n|N)$ and $P(N)$ expressions into P(n), I find that a negative factorial is inevitable but maybe I'm missing something.
 A: The partial sum (lower or upper) of the series for $e^x$ is related to the Regularized Incomplete Gamma function
$$
Q\left( {n,z} \right)\quad \left| {\;0 \le n \in Z} \right.\;\;
 = {{\Gamma \left( {n,z} \right)} \over {\Gamma \left( n \right)}}\quad
  = e^{\,\, - z} \sum\limits_{0\, \le \,k\, \le \,n - 1} {{{z^{\,k} } \over {k!\,}}} 
$$
Coming to your problem, we have
$$
P(n\;\left| N \right.) = \left( \matrix{  N \cr   n \cr}  \right)
\eta ^{\,n} \left( {1 - \eta } \right)^{\,N - n} 
$$
which is clearly the probability of $n$ successes in $N$ trials.
We assume $N$ to follow a Poisson distribution with average $\overline N $
$$
P(N) = e^{\, - \,\overline N } {{\overline N ^{\,N} } \over {N!}}
$$
Now we have
$$
P(n\;\left| N \right.) = {{P\left( {n \wedge N} \right)} \over {P(N)}}
$$
and as rightly hinted in the text, for $P(n)$ we will have
$$
\eqalign{
  & P(n) = \sum\limits_{N = 0}^\infty  {P\left( {n \wedge N} \right)}
  = \sum\limits_{N = 0}^\infty  {P(n\;\left| N \right.)P(N)}  =   \cr 
  &  = \sum\limits_{N = 0}^\infty
  {\left( \matrix{  N \cr   n \cr}  \right)\eta ^{\,n} \left( {1 - \eta } \right)^{\,N - n}
 e^{\, - \,\overline N } {{\overline N ^{\,N} } \over {N!}}}  =   \cr 
  &  = \eta ^{\,n} \left( {1 - \eta } \right)^{\, - n} e^{\, - \,\overline N }
 \sum\limits_{N = 0}^\infty  {\left( \matrix{  N \cr   n \cr}  \right)
{{\left( {\overline N \left( {1 - \eta } \right)} \right)^{\,N} } \over {N!}}}  =   \cr 
  &  = \eta ^{\,n} \left( {1 - \eta } \right)^{\, - n} e^{\, - \,\overline N }
 \sum\limits_{N = n}^\infty  {\left( \matrix{  N \cr   n \cr}  \right)
{{\left( {\overline N \left( {1 - \eta } \right)} \right)^{\,N} } \over {N!}}}  =   \cr 
  &  = {{\eta ^{\,n} \left( {1 - \eta } \right)^{\, - n} e^{\, - \,\overline N } } \over {n!}}
\sum\limits_{N = n}^\infty  {{{\left( {\overline N \left( {1 - \eta } \right)} \right)^{\,N} }
 \over {\left( {N - n} \right)!}}}  =   \cr 
  &  = {{\eta ^{\,n} \left( {1 - \eta } \right)^{\, - n} e^{\, - \,\overline N } } \over {n!}}
\sum\limits_{N - n = 0}^\infty  {{{\left( {\overline N \left( {1 - \eta } \right)} \right)^{\,N - n + n} }
 \over {\left( {N - n} \right)!}}}  =   \cr 
  &  = {{\eta ^{\,n} \left( {1 - \eta } \right)^{\, - n}
 \left( {\overline N \left( {1 - \eta } \right)} \right)^{\,n} } \over {n!}}e^{\, - \,\overline N }
 \sum\limits_{N - n = 0}^\infty  {{{\left( {\overline N \left( {1 - \eta } \right)} \right)^{\,N - n} }
 \over {\left( {N - n} \right)!}}}  =   \cr 
  &  = {{\left( {\eta \overline N } \right)^{\,n} } \over {n!}}e^{\, - \,\overline N }
 \sum\limits_{k = 0}^\infty  {{{\left( {\overline N \left( {1 - \eta } \right)} \right)^k } \over {k!}}}  =   \cr 
  &  = {{\left( {\eta \overline N } \right)^{\,n} } \over {n!}}e^{\, - \,\overline N }
 e^{\,\overline N \left( {1 - \eta } \right)}  =   \cr 
  &  = {{\left( {\eta \overline N } \right)^{\,n} } \over {n!}}
e^{\, - \left( {\eta \,\overline N } \right)}  \cr} 
$$
That is, the key passage you got astray is at
$$
\eqalign{
  & \sum\limits_{N = n}^\infty  {{{\left( {\overline N \left( {1 - \eta } \right)} \right)^{\,N} }
 \over {\left( {N - n} \right)!}}} \; \Rightarrow
 \sum\limits_{N - n = 0}^\infty  {{{\left( {\overline N \left( {1 - \eta } \right)} \right)^{\,N - n + n} }
 \over {\left( {N - n} \right)!}}}  \Rightarrow   \cr 
  &  \Rightarrow \left( {\overline N \left( {1 - \eta } \right)} \right)^{\,n}
 \sum\limits_{k = 0}^\infty 
 {{{\left( {\overline N \left( {1 - \eta } \right)} \right)^{\,k} } \over {k!}}}  \cr} 
$$
But it is good in any case that you learned about the occurring of $Q$ !
