The Zeta function $\zeta(s)$ is defined as following $$ \zeta(s)=1+\frac1{2^s}+\frac1{3^s}+\frac1{4^s}+\frac1{5^s}+\cdots=\sum_{n=1}\frac1{n^s} $$ Now it has been shown that $$ \tag1 \zeta(s)=\frac{\eta(s)}{1-\frac 1{2^s}\cdot 2} $$ $$ \eta(s)=1-\frac1{2^s}+\frac1{3^s}-\frac1{4^s}+\frac1{5^s}- \cdots=\sum_{n=1}(-1)^n\left(-\frac1{n^s}\right) $$ Now this is my first question: Is the following true?
$$ \tag2 \eta(s)\ + 2\cdot \left(\frac1{2^s}+\frac1{4^s}+\frac1{6^s}+\frac1{8^s}+\cdots \right)= \zeta (s) $$
And so can we equate $(1)$ and $(2)$? It would look something like $$ \eta(s)\ + 2\cdot \left(1+\frac1{2^s}+\frac1{4^s}+\frac1{6^s}+\frac1{8^s}+\cdots \right)=\frac{\eta(s)}{1-\frac 1{2^s}\cdot 2} $$