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Question Number 178793 by infinityaction last updated on 21/Oct/22

$$\mathrm{find}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{\left(3n\right)!}=1+\frac{1}{3!}+\frac{1}{6!}+\frac{1}{9!}+...$$

Answered by aleks041103 last updated on 21/Oct/22

$$w=e^{\frac{2\pi i}{3}}\Rightarrow w^2=w^{\ast },w^3=1$$$$e^w=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{\left(3n\right)!}+\frac{w}{(3n+1)!}+\frac{w^2}{(3n+2)!}=S_0+{wS}_1+w^2{S}_2$$$$e^{w^2}=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{\left(3n\right)!}+\frac{w^2}{(3n+1)!}+\frac{w}{(3n+2)!}=S_0+w^2{S}_1+{wS}_2$$$$e^1=S_0+S_1+S_2$$$$e^w+e^{w^2}-2e=(w+w^2-2)(S_1+S_2)=-3(S_1+S_2)$$$$\Rightarrow S_0=e-(S_1+S_2)=e+\frac{(exp\left(w\right)+exp\left(w^2\right)-2e)}{3}=$$$$=\frac{e^1+e^w+e^{w^2}}{3}=S_0$$$$w=e^{\frac{2\pi i}{3}}=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$$$$w^2=e^{-\frac{2\pi i}{3}}=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$$$$\Rightarrow e^w=\frac{1}{\sqrt{e}}{e}^{i\sqrt{3}/2},e^{w^2}=\frac{1}{\sqrt{e}}{e}^{-i\sqrt{3}/2}$$$$\Rightarrow e^w+e^{w^2}=\frac{2}{\sqrt{e}}cos\left(\frac{\sqrt{3}}{2}\right)$$$$\Rightarrow \underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{\left(3n\right)!}=\frac{e}{3}+\frac{2}{3\sqrt{e}}cos\left(\sqrt{3}/2\right)$$

Commented by aleks041103 last updated on 21/Oct/22

Commented by Tawa11 last updated on 22/Oct/22

$$\mathrm{Great}\mathrm{sir}$$

Commented by infinityaction last updated on 22/Oct/22

$$thankssir$$

Answered by ARUNG_Brandon_MBU last updated on 22/Oct/22

$$S=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{\left(3n\right)!}=1+\frac{1}{3!}+\frac{1}{6!}+\frac{1}{9!}+\cdot \cdot \cdot$$$$=(1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+\frac{1}{6!}+\frac{1}{7!}+\frac{1}{8!}+\frac{1}{9!}+\cdot \cdot \cdot )-(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{4!}+\frac{1}{5!}+\frac{1}{7!}+\frac{1}{8!}+\cdot \cdot \cdot )$$$$=e-(\frac{1}{1!}+\frac{1}{4!}+\frac{1}{7!}+\cdot \cdot \cdot )-(\frac{1}{2!}+\frac{1}{5!}+\frac{1}{8!}+\cdot \cdot \cdot )=e-\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{(3n+1)!}-\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{(3n+2)!}$$$$\Rightarrow e=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{\left(3n\right)!}+\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{(3n+1)!}+\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{(3n+2)!}$$$$e^x=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^{3n}}{\left(3n\right)!}+\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^{3n+1}}{(3n+1)!}+\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^{3n+2}}{(3n+2)}\Rightarrow e^x=f\left(x\right)+f{}^{\prime }\left(x\right)+f{}^{″}\left(x\right)$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^n}{\left(3n\right)!}\mathrm{is}\mathrm{solution}\mathrm{to}\mathrm{this}\mathrm{equation}\mathrm{with}f\left(0\right)=1\mathrm{and}f{}^{\prime }\left(0\right)=0$$$$y_{gh}:r^2+r+1=0\Rightarrow r=-\frac{1}{2}\pm i\frac{\sqrt{3}}{2}\Rightarrow y_{gh}=e^{-\frac{x}{2}}(\alpha \cos \left(\frac{\sqrt{3}}{2}x\right)+\beta \sin \left(\frac{\sqrt{3}}{2}x\right))$$$$y_p={ke}^x=y_p^{\prime }=y_p^{″}\Rightarrow e^x={ke}^x+{ke}^x+{ke}^x\Rightarrow k=\frac{1}{3}$$$$f\left(x\right)=y_{gh}+y_p=e^{-\frac{x}{2}}(\alpha \cos \left(\frac{\sqrt{3}}{2}x\right)+\beta \sin \left(\frac{\sqrt{3}}{2}x\right))+\frac{e^x}{3},f\left(0\right)=1\Rightarrow \alpha +\frac{1}{3}=1\Rightarrow \alpha =\frac{2}{3}$$$$f{}^{\prime }\left(x\right)=e^{-\frac{x}{2}}(\frac{\sqrt{3}}{2}\beta \cos \left(\frac{\sqrt{3}}{2}x\right)-\frac{\sqrt{3}}{2}\alpha \sin \left(\frac{\sqrt{3}}{2}x\right))-\frac{1}{2}{e}^{-\frac{x}{2}}(\alpha \cos \left(\frac{\sqrt{3}}{2}x\right)+\beta \sin \left(\frac{\sqrt{3}}{2}x\right))+\frac{e^x}{3}$$$$f{}^{\prime }\left(0\right)=0\Rightarrow \frac{\sqrt{3}}{2}\beta -\frac{\alpha }{2}+\frac{1}{3}=\frac{\sqrt{3}}{2}\beta =0\Rightarrow \beta =0\Rightarrow f\left(x\right)=e^{-\frac{x}{2}}\left(\frac{2}{3}\cos \left(\frac{\sqrt{3}}{2}x\right)\right)+\frac{e^x}{3}$$$$\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{\left(3n\right)!}=f\left(1\right)=\frac{2}{3\sqrt{e}}\cos \left(\frac{\sqrt{3}}{2}\right)+\frac{e}{3}=\frac{1}{3}(e+\frac{2}{\sqrt{e}}\cos \left(\frac{\sqrt{3}}{2}\right))$$

Commented by infinityaction last updated on 22/Oct/22

$$thanks$$

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