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

help Σ_(n=1) ^∞ (1/((2n−1)!))

$$\mathrm{help} \\ $$$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2n}−\mathrm{1}\right)!} \\ $$

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

or sin h(1)

$${or}\:{sin}\:{h}\left(\mathrm{1}\right) \\ $$

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

Commented by Mikael_786 last updated on 04/Aug/20

thanks Sir

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

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

we have e^x =Σ_(n=0) ^∞ (x^n /(n!)) and e^(−x) =Σ_(n=0) ^∞ (((−1)^n x^n )/(n!)) ⇒ e^x −e^(−x) =Σ_(n=0) ^∞ (1/(n!))(1−(−1)^n )x^n =Σ_(n=0) ^∞ ((2 x^(2n+1) )/((2n+1)!)) ⇒ ((e^x −e^(−x) )/2) =Σ_(n=0) ^∞ (x^(2n+1) /((2n+1)!)) =_(n=p−1) Σ_(p=1) ^∞ (x^(2p−1) /((2p−1)!)) (=sh(x)) x =1 we get ((e−e^(−1) )/2) =Σ_(p=1) ^∞ (1/((2p−1)!)) ⇒ Σ_(p=1) ^∞ (1/((2p−1)!)) =((e−(1/e))/2) =((e^2 −1)/(2e)) ( =sh(1))

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

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