# Summation - e^x

What is the value of the summation $$\sum_{x = 1}^7 \frac{4^x}{x!}$$

I know that it has something to do with $e^x$, but that only happens when $x$ is from - to infinite. Thanks for the help.

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There are seven terms and no free variables. Just write out the terms and do the arithmetic. You end up with some definite rational number. – Henning Makholm Oct 10 '11 at 2:23
Beware, this may look like like $e^x$ at first sight but it's not ($e^x$ is defined as an infinite sum, meaning it involves limits). Here it's just a sum of seven numbers ($4^1/1!, 4^2/2!, \ldots, 4^7/7!$), nothing fancy. – Joel Cohen Oct 10 '11 at 2:24
I am looking for a more general way of doing this if it exists so I can do it for larger values. – icobes Oct 10 '11 at 2:24
Use a computer with an infinite-precision arithmetic package and a for loop, then? – Henning Makholm Oct 10 '11 at 2:27
There isn't a simpler formula than what you already have for a finite upper limit. The "closed form" involves what is called an "incomplete gamma function"; see this for instance. – J. M. Oct 10 '11 at 2:37

$\frac{16004}{315}.$ And what?
In this context (which is not the same as the context of the question), one might mention that, for every $n\geqslant3$, $$\mathrm e^4-1-r_{n+1}u_n\leqslant\sum_{x=1}^n\frac{4^x}{x!}\leqslant\mathrm e^4-1-r_{n+1}v_n,$$ with $$r_n=\frac{4^{n}}{n!},\quad u_n=\frac{n+2}{n-2},\quad v_n=1,$$ and that $u_n\to1$, $v_n\to1$ and $r_n\to0$ when $n\to\infty$.