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I read the solution of one of my exercises in discrete probability theory, and there was one line that I don't understand.

$$Pr[E] = \sum _{k=1}^n \frac{(-1)^{k+1}}{k!} = 1 - \sum _{k=0}^n \frac{(-1)^{k}}{k!}$$

I've written out the sums and they are equal. However I wonder how to derive the result from the first sum.

I found this answer, on how to change the start index of a sum. However if I understood the method correctly, I end up with

$$Pr[E] = \sum _{k=1}^n \frac{(-1)^{k+1}}{k!} = \sum _{k=0}^n \frac{(-1)^{k+2}}{(k+1)!} = \sum _{k=0}^n \frac{(-1)^{k}}{(k+1)!}$$

which is also correct. The problem is: I had to calculate the upper result to use the exponential series

$$e^x = \sum_{k=0}^{\infty} \frac{x^n}{n!}$$

in order to end up with an estimation of a probability.

How do I derive

$$1 - \sum _{k=0}^n \frac{(-1)^{k}}{k!}$$ ?

Please excuse this question if it is trivial, but I tried and cannot figure this out.

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There are a few steps here.

First change the exponent on $(-1)$ as in $ \sum _{k=1}^n \frac{(-1)^{k+1}}{k!} = - \sum _{k=1}^n \frac{(-1)^{k}}{k!}$.

Second, since we want to start the index from $k=0$, we can do so, but we need to subtract the corresponding value so we haven't changed the sum. Look at it this way: $- \sum _{k=1}^n \frac{(-1)^{k}}{k!} = - ( (\sum _{k=0}^n \frac{(-1)^{k}}{k!}) - (\sum _{k=0}^0 \frac{(-1)^{k}}{k!}) = - ( (\sum _{k=0}^n \frac{(-1)^{k}}{k!}) - 1)$.

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  • $\begingroup$ I think subtracting the term of k=0 is what I couldn't figure out. This is very helpful for other sums, where you have to change the index. Thank you very much! $\endgroup$
    – Joliver
    Oct 2 '17 at 14:54
  • $\begingroup$ @Joliver: Delighted to help! $\endgroup$
    – copper.hat
    Oct 2 '17 at 14:55

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