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Question Number 100468 by Mikael_786 last updated on 26/Jun/20

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

Answered by mathmax by abdo last updated on 26/Jun/20

$$\mathrm{S}=\frac{1}{2}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{2\mathrm{n}+1-1}{(2\mathrm{n}+1)!}=\frac{1}{2}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{\left(2\mathrm{n}\right)!}-\frac{1}{2}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{(2\mathrm{n}+1)!}$$$$\mathrm{we}\mathrm{have}{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}!}+\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}!}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{\mathrm{n}!}(1+{(-1)}^{\mathrm{n}}){\mathrm{x}}^{\mathrm{n}}$$$$=2\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\mathrm{x}}^{2\mathrm{n}}}{\left(2\mathrm{n}\right)!}\Rightarrow \frac{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}}{2}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\mathrm{x}}^{2\mathrm{n}}}{\left(2\mathrm{n}\right)!}\Rightarrow \sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{\left(2\mathrm{n}\right)!}=\frac{\mathrm{e}+{\mathrm{e}}^{-1}}{2}$$$${\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{-\mathrm{x}}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{\mathrm{n}!}(1-{(-1)}^{\mathrm{n}}){\mathrm{x}}^{\mathrm{n}}=2\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\mathrm{x}}^{2\mathrm{n}+1}}{(2\mathrm{n}+1)!}\Rightarrow$$$$\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\mathrm{x}}^{2\mathrm{n}+1}}{(2\mathrm{n}+1)!}=\frac{{\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{-\mathrm{x}}}{2}\Rightarrow \sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{(2\mathrm{n}+1)!}=\frac{\mathrm{e}-{\mathrm{e}}^{-1}}{2}\Rightarrow$$$$\mathrm{S}=\frac{1}{4}(\mathrm{e}+{\mathrm{e}}^{-1})-\frac{1}{4}(\mathrm{e}-{\mathrm{e}}^{-1})=\frac{1}{4}(\mathrm{e}+{\mathrm{e}}^{-1}-\mathrm{e}+{\mathrm{e}}^{-1})=\frac{{2\mathrm{e}}^{-1}}{4}=\frac{1}{2\mathrm{e}}$$$$\mathrm{S}=\frac{1}{2\mathrm{e}}$$

Commented by Mikael_786 last updated on 27/Jun/20

$$thankyouSir$$

Answered by smridha last updated on 26/Jun/20

$$\mathit{a}\mathit{n}\mathit{s}:\frac{1}{2\mathit{e}}.\mathit{a}\mathit{p}\mathit{p}\mathit{l}\mathit{y}\mathit{g}\mathit{e}\mathit{n}\mathit{e}\mathit{r}\mathit{a}\mathit{l}\mathit{i}\mathit{s}\mathit{e}\mathit{d}$$$$\mathit{h}\mathit{y}\mathit{p}\mathit{e}\mathit{r}\mathit{g}\mathit{e}\mathit{o}\mathit{m}\mathit{e}\mathit{t}\mathit{r}\mathit{i}\mathit{c}{\mathit{f}}^{\mathit{n}}$$

Answered by maths mind last updated on 26/Jun/20

$$f\left(x\right)=\sum _{k⩾1}\frac{x^{2n+1}}{(2n+1)!}=\frac{e^x-e^{-x}}{2}$$$$\Rightarrow \frac{e^x-e^{-x}}{2x}=\sum _{n⩾1}\frac{x^{2n}}{(2n+1)!}$$$$\partial x\frac{1}{2}\left(\frac{e^x-e^{-x}}{x}\right){\mid }_{x=1}=\sum _{k⩾1}\frac{2n}{(2n+1)!}$$$$\iff \frac{1}{z}(\frac{e^x+e^{-x}}{x}-\frac{e^x-e^{-x}}{x^2}){\mid }_{x=1}=2\underset{n=1}{\sum ^{+\mathrm{\infty }}}\frac{n}{(2n+1)!}$$$$\iff e^{-1}=2\mathrm{\Sigma }\frac{n}{(2n+1)!}\Rightarrow \frac{1}{2e}=\sum _{n⩾1}\frac{n}{(2n+1)!}$$

Commented by Mikael_786 last updated on 27/Jun/20

$$thankyouSir$$

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