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.


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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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