0
$\begingroup$

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.

  • 0
    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.2011-10-10
  • 0
    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.2011-10-10
  • 0
    I am looking for a more general way of doing this if it exists so I can do it for larger values.2011-10-10
  • 0
    Use a computer with an infinite-precision arithmetic package and a for loop, then?2011-10-10
  • 0
    I doubt there is a simpler formula. One way to efficiently compute would be to write as $1 + 4 (1 + \frac{4}{2}(1+\frac{4}{3} (1 + \ldots (1 + \frac{4}{7})\ldots)$. But the more terms you put in the sum, the closer it gets to $e^4$. So at some point $e^4$ will give an accurate approximation.2011-10-10
  • 5
    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](http://math.stackexchange.com/questions/50746) for instance.2011-10-10
  • 0
    That $x$ in the summation makes my hands itch.2011-10-10

1 Answers 1

5

$\frac{16004}{315}.$ 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 $n\to\infty$.