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Question Number 139457 by mnjuly1970 last updated on 27/Apr/21

_ prove that :: Σ_(n=0) ^∞ (1/((3n)!)) =(e/3)+(2/(3(√e))) cos(((√3)/2))

$$\:_{} \:\:\: \\ $$$$\:\:\:\:\:{prove}\:{that}\::: \\ $$$$\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{3}{n}\right)!}\:=\frac{{e}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}\sqrt{{e}}}\:{cos}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$$ \\ $$

Answered by Dwaipayan Shikari last updated on 27/Apr/21

Σ_(n=0) ^∞ (x^(3n) /((3n)!))=((e^x +e^(ωx) +e^(ω^2 x) )/3) ω=e^(2πi/3) =((e^x +2e^(−(x/2)) cos(((√3)/2)x))/3) x=1:→Σ_(n=0) ^∞ (1/((3n)!))=((e+2e^(−(1/2)) cos(((√3)/2)))/3)

$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{3}{n}} }{\left(\mathrm{3}{n}\right)!}=\frac{{e}^{{x}} +{e}^{\omega{x}} +{e}^{\omega^{\mathrm{2}} {x}} }{\mathrm{3}}\:\:\omega={e}^{\mathrm{2}\pi{i}/\mathrm{3}} \\ $$$$=\frac{{e}^{{x}} +\mathrm{2}{e}^{−\frac{{x}}{\mathrm{2}}} {cos}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{x}\right)}{\mathrm{3}}\: \\ $$$${x}=\mathrm{1}:\rightarrow\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{3}{n}\right)!}=\frac{{e}+\mathrm{2}{e}^{−\frac{\mathrm{1}}{\mathrm{2}}} {cos}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)}{\mathrm{3}} \\ $$

Commented by Dwaipayan Shikari last updated on 27/Apr/21

For n=4 it is Σ_(n=0) ^∞ (x^(4n) /((4n)!))=((Σ_(i=1) ^(Roots of x^4 =1) e^(a_1 x) )/4)=((e^(−ix) +e^(ix) +e^x +e^(−x) )/4) =((cosh(x)+cos(x))/2) For n=2 Σ_(n=0) ^∞ (x^(2n) /((2n)!))=((Σ_(i=1) ^(Roots of x^2 =1) e^(a_i x) )/2)=((e^x +e^(−x) )/2)=cosh(x) For n=1 Σ_(n=0) ^∞ (x^n /(n!))=((Σ_(i=1) ^(Roots of x=1) e^(a_i x) )/1) =e^x

$${For}\:{n}=\mathrm{4} \\ $$$${it}\:{is}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{4}{n}} }{\left(\mathrm{4}{n}\right)!}=\frac{\underset{{i}=\mathrm{1}} {\overset{{Roots}\:{of}\:{x}^{\mathrm{4}} =\mathrm{1}} {\sum}}{e}^{{a}_{\mathrm{1}} {x}} }{\mathrm{4}}=\frac{{e}^{−{ix}} +{e}^{{ix}} +{e}^{{x}} +{e}^{−{x}} }{\mathrm{4}} \\ $$$$=\frac{{cosh}\left({x}\right)+{cos}\left({x}\right)}{\mathrm{2}} \\ $$$${For}\:{n}=\mathrm{2} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!}=\frac{\underset{{i}=\mathrm{1}} {\overset{{Roots}\:{of}\:{x}^{\mathrm{2}} =\mathrm{1}} {\sum}}{e}^{{a}_{{i}} {x}} }{\mathrm{2}}=\frac{{e}^{{x}} +{e}^{−{x}} }{\mathrm{2}}={cosh}\left({x}\right) \\ $$$${For}\:\:{n}=\mathrm{1} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}!}=\frac{\underset{{i}=\mathrm{1}} {\overset{{Roots}\:{of}\:{x}=\mathrm{1}} {\sum}}{e}^{{a}_{{i}} {x}} }{\mathrm{1}}\:={e}^{{x}} \\ $$

Commented by mnjuly1970 last updated on 27/Apr/21

very nice...

$$\:{very}\:{nice}... \\ $$

Commented by Dwaipayan Shikari last updated on 27/Apr/21

But is it true for Σ_(n=0) ^∞ (x^(an) /((an)!))=((Σ_(i=1) ^(Roots of x^a =1) e^(a_i x) )/a) ?? Any idea sir?

$${But}\:{is}\:{it}\:{true}\:{for} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{{an}} }{\left({an}\right)!}=\frac{\underset{{i}=\mathrm{1}} {\overset{{Roots}\:{of}\:{x}^{{a}} =\mathrm{1}} {\sum}}{e}^{{a}_{{i}} {x}} }{{a}}\:??\:\:{Any}\:{idea}\:{sir}? \\ $$

Commented by Dwaipayan Shikari last updated on 27/Apr/21

If i go for Σ_(n=0) ^∞ (x^(6n) /((6n)!))=((Σ_(i=1) ^(Roots of x^6 =1) e^(a_i x) )/6)=((e^x +e^(−x) +e^((1/2)x) (e^(((√3)/2)ix) +e^(−((√3)/2)ix) )+e^(−(x/2)) (e^(((√3)/2)ix) +e^(−((√3)/2)ix) ))/6) =((e^x +e^(−x) +2cos(((√3)/2)x)(e^(x/2) +e^(−x/2) ))/6)=((cosh(x)+2cos(((√3)/2)x)cosh((x/2)))/3) ???

$${If}\:{i}\:{go}\:{for} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{6}{n}} }{\left(\mathrm{6}{n}\right)!}=\frac{\underset{{i}=\mathrm{1}} {\overset{{Roots}\:{of}\:{x}^{\mathrm{6}} =\mathrm{1}} {\sum}}{e}^{{a}_{{i}} {x}} }{\mathrm{6}}=\frac{{e}^{{x}} +{e}^{−{x}} +{e}^{\frac{\mathrm{1}}{\mathrm{2}}{x}} \left({e}^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{ix}} +{e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{ix}} \right)+{e}^{−\frac{{x}}{\mathrm{2}}} \left({e}^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{ix}} +{e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{ix}} \right)}{\mathrm{6}}\: \\ $$$$=\frac{{e}^{{x}} +{e}^{−{x}} +\mathrm{2}{cos}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{x}\right)\left({e}^{{x}/\mathrm{2}} +{e}^{−{x}/\mathrm{2}} \right)}{\mathrm{6}}=\frac{{cosh}\left({x}\right)+\mathrm{2}{cos}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{x}\right){cosh}\left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{3}} \\ $$$$??? \\ $$

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