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Could you help me count this sum:

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

I don't think I can use binomial coefficients.

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if you expect the answer as a decimal or a fraction, just use a calculator. If you are looking for the answer in any other form, please specify. –  Ittay Weiss Jan 27 '13 at 7:57
2  
Just do it? ${}{}{}$ –  mixedmath Jan 27 '13 at 7:58
    
Well, when I am at home, of course I can use a calculator, but what if I'm at school at I need to calculate a similar sum for larger $n$s? –  Hagrid Jan 27 '13 at 8:01
    
@Hagrid see my answer for large $n$'s. –  Rustyn Jan 27 '13 at 8:06

3 Answers 3

up vote 5 down vote accepted

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.

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And the error of this approximation is bounded above by $1/(k\cdot k!)$. –  Rahul Jan 27 '13 at 9:19

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

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Well, when I am at home, of course I can use a calculator, but what if I'm at school at I need to calculate a similar sum for larger ns?

Butchered quote (ref.):

Some people, when confronted with a problem, think “I know, I'll use an approximation.” Now they have two problems.

If computing is not possible, another approach is to make a definition, e.g. define $$\varphi=\sum_{n=1}^{9} \frac{1}{n!},$$ then work with $\varphi$ instead. With this method, we don't need to account for error in the subsequent calculations (as there is none).

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