Q:

Approximations for the partial sums of exponential series

Though the question here (Partial sums of exponential series - Stack Exchange) is similar, it is more specialized and I rather need a general approximation for an arbitrary partial sum. 
Essentially, I am trying to approximate the probability mass function of a particular random variable and I ended up with a Poisson random variable's CDF in the mix. Hence, for my purpose, I need to figure out a reasonable approximation of the sum: 
$\displaystyle\sum_{k = 0}^{r} \frac{\lambda^k}{k!}$ OR the tail, i.e. $\displaystyle\sum_{k = r}^{\infty} \frac{\lambda^k}{k!}$
Does someone know some approximations for this? Also, if there exist conditions for those approximations to be valid, I'd like to know them as well. 
Thanks in advance!
Addendum: There appears to be a closed form expression for such a partial sum: 
$\displaystyle\sum_{k = 0}^{r} \frac{\lambda^k}{k!} = e^\lambda \frac{\Gamma(r + 1, \lambda)}{\Gamma(r + 1)}$, 
where $\Gamma(a, x)$ is defined as: $\displaystyle \Gamma(a, x) = \int_x^\infty t^{a - 1} e^{-t} \,dt$ and $\displaystyle \Gamma(a) = \Gamma(a, 0)$.
Is there a simple closed form approximation for the Gamma function? At the end of the day, somehow or the other, I either end up with a summation sign or an integral. I just want to be able to pin down this partial sum as a numeric quantity, that is reasonably approximate. 

A:

Use Taylor's series with remainder. We know that
$$ e^\lambda = \sum_{k=0}^r \frac{\lambda^k}{k!} + \frac{e^{c\lambda}\lambda^{r+1}}{(r+1)!}, $$
for some $c \in [0,1]$. Therefore
$$ \frac{\lambda^{r+1}}{(r+1)!} \leq e^\lambda - \sum_{k=0}^r \frac{\lambda^k}{k!} \leq e^\lambda \frac{\lambda^{r+1}}{(r+1)!}. $$
You can also get these estimates using more elementary means:
$$
\sum_{k=r+1}^\infty \frac{\lambda^k}{k!} = \frac{\lambda^{r+1}}{(r+1)!} \left[ 1 + \frac{\lambda}{r+2} + \frac{\lambda^2}{(r+2)(r+3)} + \cdots \right] < \frac{\lambda^{r+1}}{(r+1)!} \sum_{t=0}^\infty \frac{\lambda^t}{t!} = \frac{\lambda^{r+1}}{(r+1)!} e^\lambda.
$$
We can get a different upper bound by comparison to a geometric series, when $\lambda < r+2$:
$$
\sum_{k=r+1}^\infty \frac{\lambda^k}{k!} \leq \frac{\lambda^{r+1}}{(r+1)!} \sum_{t=0}^\infty \left(\frac{\lambda}{r+2}\right)^t = \frac{\lambda^{r+1}}{(r+1)!} \frac{r+2}{r+2-\lambda}.
$$

