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Question Number 38195 by maxmathsup by imad last updated on 22/Jun/18
let x≥1 and δ(x)=Σ_(n=1) ^∞ (((−1)^n )/n^x ) 1) calculate δ(x) interms of ξ(x) if x>1 2)find δ(1) 3) find the value of Σ_(n=1) ^∞ (1/((2n+1)^2 )) 4) calculate δ(3) interms of ξ(3).
$${let}\:{x}\geqslant\mathrm{1}\:{and}\:\delta\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{{x}} } \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:\delta\left({x}\right)\:{interms}\:{of}\:\xi\left({x}\right)\:{if}\:{x}>\mathrm{1} \\ $$$$\left.\mathrm{2}\right){find}\:\:\delta\left(\mathrm{1}\right) \\ $$$$\left.\mathrm{3}\right)\:{find}\:{the}\:{value}\:{of}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{4}\right)\:{calculate}\:\delta\left(\mathrm{3}\right)\:{interms}\:{of}\:\xi\left(\mathrm{3}\right). \\ $$
Commented by math khazana by abdo last updated on 23/Jun/18
1) δ(x)= Σ_(n=1) ^∞ (1/((2n)^x )) −Σ_(n=0) ^∞ (1/((2n+1)^x )) = 2^(−x) ξ(x) −Σ_(n=0) ^∞ (1/((2n+1)^x )) but we have ξ(x)=Σ_(n=1) ^∞ (1/n^x ) = Σ_(n=1) ^∞ (1/((2n)^x )) +Σ_(n=0) ^∞ (1/((2n+1)^x )) =2^(−x) ξ(x) +Σ_(n=0) ^∞ (1/((2n+1)^x )) ⇒ Σ_(n=0) ^∞ (1/((2n+1)^x )) =(1−2^(−x) )ξ(x) ⇒ δ(x)= 2^(−x) ξ(x) −(1−2^(−x) )ξ(x) =(2^(1−x) −1)ξ(x) 2) δ(1) =Σ_(n=1) ^∞ (((−1)^n )/n) =−ln(2) 3)we have Σ_(n=0) ^∞ (1/((2n+1)^2 )) =(1−2^(−1) )ξ(2) =(3/4) Σ_(n=1) ^∞ (1/n^2 ) =(3/4) (π^2 /6) = (π^2 /8) Σ_(n=0) ^∞ (1/((2n+1)^2 )) =(π^2 /8) . 4)δ(3)=(2^(1−3) −1)ξ(3)= ((1/4)−1)ξ(3) =−(3/4)ξ(3) ⇒ Σ_(n=1) ^∞ (((−1)^n )/n^3 ) =−(3/4) Σ_(n=1) ^∞ (1/n^3 ) .
$$\left.\mathrm{1}\right)\:\delta\left({x}\right)=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)^{{x}} }\:−\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{{x}} } \\ $$$$=\:\mathrm{2}^{−{x}} \:\xi\left({x}\right)\:−\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{{x}} }\:{but}\:{we}\:{have} \\ $$$$\xi\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}^{{x}} }\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)^{{x}} }\:+\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{{x}} } \\ $$$$=\mathrm{2}^{−{x}} \xi\left({x}\right)\:+\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{{x}} }\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{{x}} }\:=\left(\mathrm{1}−\mathrm{2}^{−{x}} \right)\xi\left({x}\right)\:\Rightarrow \\ $$$$\delta\left({x}\right)=\:\mathrm{2}^{−{x}} \xi\left({x}\right)\:−\left(\mathrm{1}−\mathrm{2}^{−{x}} \right)\xi\left({x}\right) \\ $$$$=\left(\mathrm{2}^{\mathrm{1}−{x}} \:−\mathrm{1}\right)\xi\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:\delta\left(\mathrm{1}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:=−{ln}\left(\mathrm{2}\right) \\ $$$$\left.\mathrm{3}\right){we}\:{have}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:=\left(\mathrm{1}−\mathrm{2}^{−\mathrm{1}} \right)\xi\left(\mathrm{2}\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\frac{\mathrm{3}}{\mathrm{4}}\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:. \\ $$$$\left.\mathrm{4}\right)\delta\left(\mathrm{3}\right)=\left(\mathrm{2}^{\mathrm{1}−\mathrm{3}} −\mathrm{1}\right)\xi\left(\mathrm{3}\right)=\:\left(\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}\right)\xi\left(\mathrm{3}\right) \\ $$$$=−\frac{\mathrm{3}}{\mathrm{4}}\xi\left(\mathrm{3}\right)\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{3}} }\:=−\frac{\mathrm{3}}{\mathrm{4}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\:. \\ $$

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