Dirichlet series for inverse of Eta function We know that $$ \frac{1}{\zeta (s)} = \sum_{n=1}^{\infty} \frac{\mu(n)}{n^{s}} $$
but what happens with  $$ \frac{1}{\eta (s)} = \sum_{n=1}^{\infty} \frac{b(n)}{n^{s}} $$
with $ \eta (s) =  (1-2^{1-s})\zeta (s) ?$
Can I evaluate the coefficients $ b(n) $ ? Perhaps I should apply the mobius transform to the sequence $ a(n)= (-1)^{n} .$

$$\frac{1}{\eta(s)}=\frac1{(1-2^{1-s})\cdot\zeta(s)}=\frac1{1-2^{1-s}}\cdot\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{s}}=\sum_{n=1}^{\infty}\frac{b(n)}{n^{s}}\iff b(n)=\frac{\mu(n)}{1-2^{1-s}}$$
