Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 5 down vote accepted

$\frac{16004}{315}.$ And what?

Edit The OP wrote in a comment:

I am looking for a more general way of doing this if it exists so I can do it for larger values.

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 $n\to\infty$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.