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Question Number 166660 by mnjuly1970 last updated on 24/Feb/22

$$$$$$calculate$$$$\mathrm{\Omega }=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{\left(3n\right)!}=?$$$$$$

Answered by amin96 last updated on 24/Feb/22

$$\underset{1}{\underbrace{{\mathbf{e}}^{\mathbf{x}}=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\mathbf{x}}^{\mathbf{n}}}{\mathbf{n}!}}}\underset{2}{\underbrace{{\mathbf{e}}^{\mathbf{xk}}=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(\mathit{x}\mathit{k}\right)}^{\mathit{n}}}{\mathit{n}!}}}\underset{3}{\underbrace{{\mathit{e}}^{{\mathit{x}\mathit{k}}^2}=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\mathit{x}}^{\mathit{n}}{\mathit{k}}^{2\mathit{n}}}{\mathit{n}!}}}$$$$\left(1\right)+\left(2\right)+\left(3\right)=3\times \underset{S}{\underbrace{\left(\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^{3n}}{\left(3n\right)!}\right)}}$$$$S=\frac{e^x+e^{kx}+e^{k^{2}x}}{3}k=\frac{-1+i\sqrt{3}}{2}{k}^2=\frac{-1-i\sqrt{3}}{2}$$$$S=\frac{1}{3}\times (e^x+e^{x\times \frac{-1+i\sqrt{3}}{2}}+e^{x\times \frac{-1-i\sqrt{3}}{2}})=$$$$=\frac{1}{3}\times (e^x+e^{-\frac{x}{2}}(e^{\frac{xi\sqrt{3}}{2}}+e^{-\frac{xi\sqrt{3}}{2}}))=\frac{1}{3}\times (e^x+2e^{-\frac{x}{2}}cos\left(\frac{x\sqrt{3}}{2}\right))$$$$x=1$$$$S=\underset{\mathbf{n}=0}{\sum ^{\mathrm{\infty }}}\frac{1}{\left(3\mathbf{n}\right)!}=\frac{1}{3}\times (e+2e^{-\frac{1}{2}}cos\left(\frac{\sqrt{3}}{2}\right))\mathbf{by}\mathbf{MATH}.\mathbf{AMIN}$$

Commented by mnjuly1970 last updated on 25/Feb/22

$$bravosir$$

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