Question Number 26223 by abdo imad last updated on 22/Dec/17
$${let}\:{put}\:\xi\left({x}\right)=\:\:\sum_{{n}=\mathrm{1}} ^{\propto} \:\:\frac{\mathrm{1}}{{n}^{{x}} }\:\:{with}\:{x}>\mathrm{1} \\ $$$${and}\:\:\delta\left({x}\right)\:\:=\sum_{{n}=\mathrm{1}} ^{\propto} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{{x}} }\:\:\:{find}\:{a}\:{relation} \\ $$$${between}\:\xi\left({x}\right)\:{and}\:\delta\left({x}\right). \\ $$
Commented by abdo imad last updated on 23/Dec/17
$$\delta\left({x}\right)\:=\:\:\sum_{{p}=\mathrm{1}} ^{\propto} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}\right)^{{x}} }\:\:−\:\sum_{{p}=\mathrm{0}} ^{\propto} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{{x}} }=\:\mathrm{2}^{−{x}} \xi\left({x}\right)−\:\sum_{{p}=\mathrm{0}} ^{\propto} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{{x}} } \\ $$$${but}\:\:\xi\left({x}\right)\:\:=\:\:\sum_{{p}=\mathrm{1}} ^{\propto} \:\frac{\mathrm{1}}{\left(\mathrm{2}{p}\right)^{{x}} }\:\:+\:\sum_{{p}=\mathrm{0}} ^{\propto} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{{x}} } \\ $$$$\Rightarrow\:\:\:\sum_{{p}=\mathrm{0}} ^{\propto} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{{x}} }\:\:=\:\xi\left({x}\right)\:−\mathrm{2}^{−{x}} \xi\left({x}\right)\:=\left(\mathrm{1}−\mathrm{2}^{−{x}} \right)\xi\left({x}\right) \\ $$$$\Rightarrow\:\:\delta\left({x}\right)=\:\mathrm{2}^{−{x}} −\left(\mathrm{1}−\mathrm{2}^{−{x}} \right)\xi\left({x}\right)=\left(\:\mathrm{2}^{\mathrm{1}−{x}} \:−\mathrm{1}\right)\xi\left({x}\right) \\ $$
Answered by prakash jain last updated on 22/Dec/17
$$\delta\left({x}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{{x}} } \\ $$$$=−\left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{{x}} }−\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)^{{x}} }\right) \\ $$$$−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{{x}} }−\frac{\mathrm{1}}{\mathrm{3}^{{x}} }+..=−\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{{x}} }+\frac{\mathrm{1}}{\mathrm{3}^{{x}} }\right)..+\mathrm{2}\left[\frac{\mathrm{1}}{\mathrm{2}^{{x}} }+\frac{\mathrm{1}}{\mathrm{4}^{{x}} }+..\right] \\ $$$$=−\left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{{x}} }−\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)^{{x}} }\right) \\ $$$$\delta\left({x}\right)=−\zeta\left({x}\right)+\frac{\mathrm{2}}{\mathrm{2}^{{x}} }\zeta\left({x}\right) \\ $$$$\delta\left({x}\right)=\left(\mathrm{2}^{{x}−\mathrm{1}} −\mathrm{1}\right)\zeta\left({x}\right) \\ $$
Commented by prakash jain last updated on 22/Dec/17
$$\delta\left({x}\right)=−\eta\left({x}\right)\:\left(\mathrm{eta}\:\mathrm{function}\right) \\ $$