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Question Number 106365 by Mikael_786 last updated on 04/Aug/20

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

Commented by Dwaipayan Shikari last updated on 04/Aug/20

$$orsinh\left(1\right)$$

Answered by Dwaipayan Shikari last updated on 04/Aug/20

Commented by Mikael_786 last updated on 04/Aug/20

$$\mathrm{thanks}\mathrm{Sir}$$

Commented by Ar Brandon last updated on 04/Aug/20

��wow ! What triggered you into using this identity, Shikari ? Or it's just from experience ?��

Commented by Dwaipayan Shikari last updated on 05/Aug/20

It includes factorial series .So I assumed exponential series may be useful.��

Commented by Ar Brandon last updated on 05/Aug/20

Alright, cool �� That was brilliant ��

Answered by mathmax by abdo last updated on 04/Aug/20

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

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