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
![For s∈R, s>1 η(s)=(1/1^s )−(1/2^s )+(1/3^s )−(1/4^s )+.. η(s)=(1/1^s )+((1/2^s )−(2/2^s ))+(1/3^s )+((1/4^s )−(2/4^s ))+... taking all −ve terms towards end. series rearrangement is valid for s>1. η(s)=ζ(s)−(2/2^s )[(1/1^s )+(1/2^s )+...] η(s)=ζ(s)−2^(1−s) ζ(s) η(s)=ζ(s)(1−2^(1−s) )](https://www.tinkutara.com/question/Q2841.png)
$$\mathrm{For}\:{s}\in\mathbb{R},\:{s}>\mathrm{1} \\ $$$$\eta\left({s}\right)=\frac{\mathrm{1}}{\mathrm{1}^{{s}} }−\frac{\mathrm{1}}{\mathrm{2}^{{s}} }+\frac{\mathrm{1}}{\mathrm{3}^{{s}} }−\frac{\mathrm{1}}{\mathrm{4}^{{s}} }+.. \\ $$$$\eta\left({s}\right)=\frac{\mathrm{1}}{\mathrm{1}^{{s}} }+\left(\frac{\mathrm{1}}{\mathrm{2}^{{s}} }−\frac{\mathrm{2}}{\mathrm{2}^{{s}} }\right)+\frac{\mathrm{1}}{\mathrm{3}^{{s}} }+\left(\frac{\mathrm{1}}{\mathrm{4}^{{s}} }−\frac{\mathrm{2}}{\mathrm{4}^{{s}} }\right)+… \\ $$$${taking}\:{all}\:−{ve}\:{terms}\:{towards}\:{end}. \\ $$$${series}\:{rearrangement}\:{is}\:{valid}\:{for}\:{s}>\mathrm{1}. \\ $$$$\eta\left({s}\right)=\zeta\left({s}\right)−\frac{\mathrm{2}}{\mathrm{2}^{{s}} }\left[\frac{\mathrm{1}}{\mathrm{1}^{{s}} }+\frac{\mathrm{1}}{\mathrm{2}^{{s}} }+…\right] \\ $$$$\eta\left({s}\right)=\zeta\left({s}\right)−\mathrm{2}^{\mathrm{1}−{s}} \zeta\left({s}\right) \\ $$$$\eta\left({s}\right)=\zeta\left({s}\right)\left(\mathrm{1}−\mathrm{2}^{\mathrm{1}−{s}} \right) \\ $$