$$\sum^{n-m}_{k=0}\frac{(-1)^k}{k!\times (n-k+1)!} = $$
How to simplify this to a one-term form. I have trouble with it. Thank you for reading and trying.
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Do you have some reason to believe that what you are asking for is possible? – Gerry Myerson Oct 15 '15 at 12:10
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@GerryMyerson Optimism? – Akiva Weinberger Oct 15 '15 at 13:37
2 Answers
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$$\frac{1}{k!(n-k+1)!}=\frac{1}{n+1}\frac{(n-k+1)+(k)}{k!(n-k+1)!}=\frac{1}{n+1}(\frac{1}{k!(n-k)!}+\frac{1}{(k-1)!(n-(k-1))!})$$ so $$\frac{(-1)^k}{k!(n-k+1)!}=\frac{1}{n+1}(\frac{(-1)^k}{k!(n-k)!}-\frac{(-1)^{k-1}}{(k-1)!(n-(k-1))!})$$ Now use telescope rule.
R.N
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$$\begin{align} \sum_{k=0}^{n-m}\frac{(-1)^k}{k!(n-k+1)!} &=\sum_{k=0}^{n-m}\frac{(-1)^k}{(n+1)!}\cdot \frac{(n+1)!}{k!(n+1-k)!}\\ &=\frac 1{(n+1)!}\color{blue}{\sum_{k=0}^{n-m}(-1)^k\binom {n+1}k}\\ &=\frac 1{(n+1)!}\color{blue}{(-1)^{n-m}\binom n{n-m}}\qquad\qquad \text{(*)}\\ &=\frac{(-1)^{n-m}}{(n+1)!}\binom nm\qquad\blacksquare \end{align}$$
*refer to this.
Hypergeometricx
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