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Question Number 142655 by mnjuly1970 last updated on 03/Jun/21

$$evaluate.....$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\left(\frac{n.cos\left(n\pi \right)}{\mathrm{\Gamma }(2n+2)}\right)=?.....$$$$.......$$

Answered by qaz last updated on 03/Jun/21

$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{\mathrm{n}\cdot \cos \left(\mathrm{n}\pi \right)}{\mathrm{\Gamma }(2\mathrm{n}+2)}$$$$=\mathrm{\Re }\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{\mathrm{n}\cdot {\mathrm{e}}^{\mathrm{in}\pi }}{(2\mathrm{n}+1)!}$$$$=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{\mathrm{n}\cdot {(-1)}^{\mathrm{n}}}{(2\mathrm{n}+1)!}$$$$=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{\mathrm{n}\cdot {(-1)}^{\mathrm{n}}}{(2\mathrm{n}+1)!}{\mathrm{x}}^{2\mathrm{n}+1}{\mid }_{\mathrm{x}=1}$$$$=\frac{1}{2}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{[\frac{(2\mathrm{n}+1)}{(2\mathrm{n}+1)!}{(-1)}^{\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}+1}-\frac{{(-1)}^{\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}+1}}{(2\mathrm{n}+1)!}]}_{\mathrm{x}=1}$$$$=\frac{1}{2}(\mathrm{xDsin}\mathrm{x}-\sin \mathrm{x}){\mid }_{\mathrm{x}=1}$$$$=\frac{1}{2}(\mathrm{xcos}\mathrm{x}-\sin \mathrm{x}){\mid }_{\mathrm{x}=1}$$$$=\frac{1}{2}(\cos 1-\sin 1)$$$$\mathrm{nice}\mathrm{problem}.\mathrm{sir}$$

Commented by mnjuly1970 last updated on 03/Jun/21

$$thanksalotmrqaz$$$$veryniceasalways...$$

Answered by Dwaipayan Shikari last updated on 03/Jun/21

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n+1}n}{(2n+1)!}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n+1}{x}^{2n+1}}{(2n+1)!}=sin\left(x\right)$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n+1}{x}^n}{(2n+1)!}=\frac{sin\left(\sqrt{x}\right)}{\sqrt{x}}$$$$D.B.S.W.R.T.x{\mid }_{x=1}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n}n}{(2n+1)!}=\frac{d}{dx}{\mid }_{x=1}\frac{sin\left(\sqrt{x}\right)}{\sqrt{x}}=\frac{\frac{1}{2}cos\left(\sqrt{x}\right)-\frac{1}{2\sqrt{x}}sin\left(\sqrt{x}\right)}{x}{\mid }_{x=1}$$$$=\frac{1}{2}(cos1-sin1)$$

Commented by mnjuly1970 last updated on 03/Jun/21

$$thanksalotgrateful...$$$$Differentiationbothsideswiththerespect$$$$tox$$

Commented by Dwaipayan Shikari last updated on 03/Jun/21

$$Hahahha$$

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