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Question Number 166660 by mnjuly1970 last updated on 24/Feb/22
calculate Ω = Σ_(n=0) ^∞ (1/((3n)!)) = ?
$$ \\ $$$$\:\:\:{calculate} \\ $$$$\:\:\:\:\Omega\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{3}{n}\right)!}\:\:=\:? \\ $$$$\:\:\:\:\: \\ $$
Answered by amin96 last updated on 24/Feb/22
e^x =Σ_(n=0) ^∞ (x^n /(n!))_(1) e^(xk) =Σ_(n=0) ^∞ (((xk)^n )/(n!)) _(2) e^(xk^2 ) =Σ_(n=0) ^∞ ((x^n k^(2n) )/(n!))_(3) (1)+(2)+(3)=3×(Σ_(n=0) ^∞ (x^(3n) /((3n)!)))_(S) S=((e^x +e^(kx) +e^(k^2 x) )/3) k=((−1+i(√3))/2) k^2 =((−1−i(√3))/2) S=(1/3)×(e^x +e^(x×((−1+i(√3))/2)) +e^(x×((−1−i(√3))/2)) )= =(1/3)×(e^x +e^(−(x/2)) (e^((xi(√3))/2) +e^(−((xi(√3))/2)) ))=(1/3)×(e^x +2e^(−(x/2)) cos(((x(√3))/2))) x=1 S=Σ_(n=0) ^∞ (1/((3n)!))=(1/3)×(e+2e^(−(1/2)) cos(((√3)/2))) by MATH.AMIN
$$\underset{\mathrm{1}} {\underbrace{\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{n}}} }{\boldsymbol{\mathrm{n}}!}}}\:\:\:\underset{\mathrm{2}} {\underbrace{\:\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{xk}}} =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\boldsymbol{{xk}}\right)^{\boldsymbol{{n}}} }{\boldsymbol{{n}}!}\:}}\:\:\underset{\mathrm{3}} {\underbrace{\boldsymbol{{e}}^{\boldsymbol{{xk}}^{\mathrm{2}} } =\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\boldsymbol{{x}}^{\boldsymbol{{n}}} \boldsymbol{{k}}^{\mathrm{2}\boldsymbol{{n}}} }{\boldsymbol{{n}}!}}} \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right)+\left(\mathrm{3}\right)=\mathrm{3}×\underset{{S}} {\underbrace{\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{3}{n}} }{\left(\mathrm{3}{n}\right)!}\right)}} \\ $$$${S}=\frac{{e}^{{x}} +{e}^{{kx}} +{e}^{{k}^{\mathrm{2}} {x}} }{\mathrm{3}}\:\:\:\:{k}=\frac{−\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:\:\:{k}^{\mathrm{2}} =\frac{−\mathrm{1}−{i}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{3}}×\left({e}^{{x}} +{e}^{{x}×\frac{−\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}} +{e}^{{x}×\frac{−\mathrm{1}−{i}\sqrt{\mathrm{3}}}{\mathrm{2}}} \right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}×\left({e}^{{x}} +{e}^{−\frac{{x}}{\mathrm{2}}} \left({e}^{\frac{{xi}\sqrt{\mathrm{3}}}{\mathrm{2}}} +{e}^{−\frac{{xi}\sqrt{\mathrm{3}}}{\mathrm{2}}} \right)\right)=\frac{\mathrm{1}}{\mathrm{3}}×\left({e}^{{x}} +\mathrm{2}{e}^{−\frac{{x}}{\mathrm{2}}} {cos}\left(\frac{{x}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\right) \\ $$$${x}=\mathrm{1} \\ $$$${S}=\underset{\boldsymbol{\mathrm{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{3}\boldsymbol{\mathrm{n}}\right)!}=\frac{\mathrm{1}}{\mathrm{3}}×\left({e}+\mathrm{2}{e}^{−\frac{\mathrm{1}}{\mathrm{2}}} {cos}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\right)\:\boldsymbol{\mathrm{by}}\:\boldsymbol{\mathrm{MATH}}.\boldsymbol{\mathrm{AMIN}} \\ $$
Commented by mnjuly1970 last updated on 25/Feb/22
bravo sir
$$\:\:\:{bravo}\:{sir} \\ $$

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