Question Number 2815 by prakash jain last updated on 28/Nov/15

Commented by prakash jain last updated on 27/Nov/15

Commented by 123456 last updated on 28/Nov/15

Commented by prakash jain last updated on 28/Nov/15

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)