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Question Number 145345 by qaz last updated on 04/Jul/21

$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}}\mathrm{n}}{(2\mathrm{n}+1)!}=?$$

Answered by Olaf_Thorendsen last updated on 04/Jul/21

$$\sin x=\underset{n=0}{\sum ^n}{(-1)}^n\frac{x^{2n+1}}{(2n+1)!}$$$$\frac{\sin x}{x}=\underset{n=0}{\sum ^n}{(-1)}^n\frac{x^{2n}}{(2n+1)!}$$$$\frac{\sin \sqrt{x}}{\sqrt{x}}=\underset{n=0}{\sum ^n}{(-1)}^n\frac{x^n}{(2n+1)!}$$$$\frac{d}{dx}\left(\frac{\sin \sqrt{x}}{\sqrt{x}}\right)=\underset{n=1}{\sum ^n}{(-1)}^n\frac{{nx}^{n-1}}{(2n+1)!}$$$$\frac{\left(\frac{1}{2\sqrt{x}}\cos \sqrt{x}\right)\sqrt{x}-\frac{1}{2\sqrt{x}}\sin \sqrt{x}}{x}=\underset{n=1}{\sum ^n}{(-1)}^n\frac{{nx}^{n-1}}{(2n+1)!}$$$$\mathrm{For}x=1$$$$\frac{1}{2}(\cos \left(1\right)-\sin \left(1\right))=\underset{n=1}{\sum ^n}\frac{{(-1)}^{n}n}{(2n+1)!}$$

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