How to simplify the power series $\sum_{k=0}^{n-1}\frac{x^k}{k!}$?

Question

I want to simplify the terms above to a much more simple expression. How should I do that so that I could avoid the summation in my final expression? I tried to use taylor series but it seems like only applicable to infinite series. Any help will be greatly appreciated!

Answer 1

The expression

$$p(x,n) = \sum_{i=0}^{n-1} \frac{x^i}{i!}$$

when multiplied by $e^{-x}$ is the chance that a Poisson random variable $X$ with parameter $x$ will be strictly less than $n:$

$$\Pr(X \lt n) = \Pr(X \le n-1) = e^{-x}\sum_{i=0}^{n-1} \frac{x^i}{i!}.$$

This event can also be characterized as the chance that the time needed to observe the first $n$ events in a Poisson process exceeds $x.$ But that time is the sum of the waiting times to the first event, from the first to the second, and so on, up to the $n^{\text{th}}.$ Each of those waiting times has an exponential distribution (with parameter $x$) and they are independent (which is part of the definition of a Poisson process). Since the sum of $n$ identical independent exponential variables is a Gamma variable $Y$ with shape parameter $n,$ we obtain

$$\sum_{i=0}^{n-1} \frac{x^i}{i!} = p(x,n) = e^x\Pr(X \lt n) = e^x\Pr(Y \gt x) = \frac{e^x}{\Gamma(n)}\int_x^\infty t^{n-1}\,e^{-t}\,\mathrm{d}t.$$

(Having discovered this result, you can prove it by repeatedly integrating the right hand side by parts if you want a purely mathematical demonstration.)

In many settings, obtaining an integral expression for a function is useful because it can be analyzed using Calculus techniques and can be evaluated numerically with efficient "quadrature" algorithms. As such, this might be considered a simplified or even "closed form" evaluation of the original sum (which has arbitrary length). For example, from this integral you can obtain Taylor series around $x=\infty$ and around $n=\infty$ (the links go to the Wolfram Alpha site). Instead of Taylor series, though, consider asymptotic expansions or other approximations. (In particular, the Taylor series for $n$ doesn't look terribly useful: it tends to be poor unless $n$ is huge.)

Answer 2

The ways you have written it are already pretty much as simple as it gets. Either you use the summation notation on the left hand side, or you write out the terms as in the right hand side. Another alternative avoiding the use of the summation would be to use the exponential function's Taylor series representation, which is $$ e^x = \sum_{k=0}^\infty \frac{x^k}{k!}. $$ Using that, you could rewrite your expression as $$ \sum_{k=0}^{n-1} \frac{x^k}{k!} = e^x - O(x^n), $$ where $O(x^n)$ is an error term of degree $n$ (and higher). This is really only useful close to the origin though, as that is where the error term would disappear.

Answer 3

$$\sum_{k=0}^{n-1}\frac{x^k}{k!}=e^x\frac{ \Gamma (n,x)}{\Gamma (n)}$$ where appears the (upper) incomplete gamma function.