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Question Number 103410 by Ar Brandon last updated on 14/Jul/20

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

Answered by mathmax by abdo last updated on 15/Jul/20

$$\mathrm{we}\mathrm{have}{\mathrm{e}}^{-1}=\sum _{\mathrm{k}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}!}=\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}!}+\sum _{\mathrm{k}=\mathrm{n}+1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}!}$$$$=\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}!}+{\mathrm{R}}_{\mathrm{n}}\left(\mathrm{rest}\mathrm{at}\mathrm{the}\mathrm{serie}\mathrm{of}\mathrm{taylor}\right)\Rightarrow$$$$\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}!}=\frac{1}{\mathrm{e}}-{\mathrm{R}}_{\mathrm{n}}\mathrm{with}{\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{R}}_{\mathrm{n}}=0$$

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