# s-n-1-1-n-1-n-s-Dirichlet-eta-function-prove-that-s-1-2-1-s-s-

Question Number 2815 by prakash jain last updated on 28/Nov/15
$$\eta\left({s}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{{s}} }\:\mathrm{Dirichlet}\:\mathrm{eta}\:\mathrm{function} \\$$$$\mathrm{prove}\:\mathrm{that} \\$$$$\eta\left({s}\right)=\left(\mathrm{1}−\mathrm{2}^{\mathrm{1}−{s}} \right)\zeta\left({s}\right) \\$$
Commented by prakash jain last updated on 27/Nov/15
$$\mathrm{This}\:\mathrm{result}\:\mathrm{is}\:\mathrm{used}\:\mathrm{in}\:\mathrm{answer}\:\mathrm{to}\:\mathrm{Q2796}. \\$$
Commented by 123456 last updated on 28/Nov/15
$$\eta\left({s}\right)=\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{{s}} }\:\:\left(???\right) \\$$
Commented by prakash jain last updated on 28/Nov/15
$$\mathrm{Yes}.\:\mathrm{corrected}. \\$$
Answered by prakash jain last updated on 28/Nov/15
$$\mathrm{For}\:{s}\in\mathbb{R},\:{s}>\mathrm{1} \\$$$$\eta\left({s}\right)=\frac{\mathrm{1}}{\mathrm{1}^{{s}} }−\frac{\mathrm{1}}{\mathrm{2}^{{s}} }+\frac{\mathrm{1}}{\mathrm{3}^{{s}} }−\frac{\mathrm{1}}{\mathrm{4}^{{s}} }+.. \\$$$$\eta\left({s}\right)=\frac{\mathrm{1}}{\mathrm{1}^{{s}} }+\left(\frac{\mathrm{1}}{\mathrm{2}^{{s}} }−\frac{\mathrm{2}}{\mathrm{2}^{{s}} }\right)+\frac{\mathrm{1}}{\mathrm{3}^{{s}} }+\left(\frac{\mathrm{1}}{\mathrm{4}^{{s}} }−\frac{\mathrm{2}}{\mathrm{4}^{{s}} }\right)+… \\$$$${taking}\:{all}\:−{ve}\:{terms}\:{towards}\:{end}. \\$$$${series}\:{rearrangement}\:{is}\:{valid}\:{for}\:{s}>\mathrm{1}. \\$$$$\eta\left({s}\right)=\zeta\left({s}\right)−\frac{\mathrm{2}}{\mathrm{2}^{{s}} }\left[\frac{\mathrm{1}}{\mathrm{1}^{{s}} }+\frac{\mathrm{1}}{\mathrm{2}^{{s}} }+…\right] \\$$$$\eta\left({s}\right)=\zeta\left({s}\right)−\mathrm{2}^{\mathrm{1}−{s}} \zeta\left({s}\right) \\$$$$\eta\left({s}\right)=\zeta\left({s}\right)\left(\mathrm{1}−\mathrm{2}^{\mathrm{1}−{s}} \right) \\$$