Sum of reciprocals of factorials

Question

Could you help me count this sum:

$$ \sum_{n=1}^{9} \frac{1}{n!} $$

I don't think I can use binomial coefficients.

Answer 1

Note: $\sum_{n=1}^{\infty} \frac{1}{n!} = e-1$

So you can gain an approximation to: $$ \sum_{n=1}^{k} \frac{1}{n!} $$ by using $e-1$, where $k>>9$

After a short amount of terms, the difference between the sum and $e-1$ becomes negligible.

Also in general,
$$ \sum_{n=1}^k \frac{1}{n!} = e \frac{\Gamma(k+1,1)}{k!} - 1 $$ Since $k+1$ is a positive integer, we have that: $$ \Gamma(k+1,1) = \frac{k!}{e}\sum_{m=0}^{k} \frac{1}{m!} $$ Which merely shows that the incomplete gamma function serves.
The incomplete gamma function is defined by: $$ P(a,x) = \frac{1}{\Gamma(a)}\int_{0}^{x} e^{-t}t^{a-1}dt $$ This integral is non-elementary, but at school, you can use MATLAB, Mathematica, or even wolfram alpha, to evaluate this integral numerically with the objective of evaluating larger sums. But then again, you can just evaluate the sum directly with any of these tools.

Answer 2

$$\sum_{n=1}^k \frac{1}{n!} = e \frac{\Gamma(k+1,1)}{k!} - 1$$