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I have been trying to proof this equation $$\sum_{i\le s\le j} \frac{(-1)^{s-i}}{(s-i)!(j-s)!}=\delta_{ij}$$ for all $i$,$j$ natural, $i \le j$, where $\delta_{ij}$ is the Kronecker delta.

For $i=j$ it is equal to one. If $i<j$, there are two cases: either there is an even number of summands or an odd number of them. In both cases the sum is (except for the signs) 'the same', if you read it left to right or right to left and the signs go $+-+-...,$ so if there is an even number of summands you just subtract the first and the last, the second and the last but one ad so on and the result is zero. However, I do not know how to prove it if there is an odd number of summands. Can anyone help me, please? Thank you.

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Make the substitions $n = j - i, k = s - i$. Then the sum becomes $$\sum_{k=0}^n \frac{1}{n!}\binom{n}{k}(-1)^k = \frac{1}{n!}(1 - 1)^n = \begin{cases}1& n = 0\\0&n\ne 0\end{cases}$$

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