Question Number 106365 by Mikael_786 last updated on 04/Aug/20
$$\mathrm{help} \\ $$$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2n}−\mathrm{1}\right)!} \\ $$
Commented by Dwaipayan Shikari last updated on 04/Aug/20
$${or}\:{sin}\:{h}\left(\mathrm{1}\right) \\ $$
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
$$\mathrm{we}\:\mathrm{have}\:\mathrm{e}^{\mathrm{x}} \:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{x}^{\mathrm{n}} }{\mathrm{n}!}\:\mathrm{and}\:\mathrm{e}^{−\mathrm{x}} \:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{x}^{\mathrm{n}} }{\mathrm{n}!}\:\Rightarrow \\ $$$$\mathrm{e}^{\mathrm{x}} −\mathrm{e}^{−\mathrm{x}} \:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{n}!}\left(\mathrm{1}−\left(−\mathrm{1}\right)^{\mathrm{n}} \right)\mathrm{x}^{\mathrm{n}} \:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{2}\:\mathrm{x}^{\mathrm{2n}+\mathrm{1}} }{\left(\mathrm{2n}+\mathrm{1}\right)!}\:\Rightarrow \\ $$$$\frac{\mathrm{e}^{\mathrm{x}} −\mathrm{e}^{−\mathrm{x}} }{\mathrm{2}}\:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{x}^{\mathrm{2n}+\mathrm{1}} }{\left(\mathrm{2n}+\mathrm{1}\right)!}\:=_{\mathrm{n}=\mathrm{p}−\mathrm{1}} \:\:\sum_{\mathrm{p}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{x}^{\mathrm{2p}−\mathrm{1}} }{\left(\mathrm{2p}−\mathrm{1}\right)!}\:\:\left(=\mathrm{sh}\left(\mathrm{x}\right)\right) \\ $$$$\mathrm{x}\:=\mathrm{1}\:\mathrm{we}\:\mathrm{get}\:\frac{\mathrm{e}−\mathrm{e}^{−\mathrm{1}} }{\mathrm{2}}\:=\sum_{\mathrm{p}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2p}−\mathrm{1}\right)!}\:\Rightarrow \\ $$$$\sum_{\mathrm{p}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2p}−\mathrm{1}\right)!}\:=\frac{\mathrm{e}−\frac{\mathrm{1}}{\mathrm{e}}}{\mathrm{2}}\:=\frac{\mathrm{e}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2e}}\:\left(\:=\mathrm{sh}\left(\mathrm{1}\right)\right)\:\: \\ $$