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} .$
 A: We can find the coefficients $b(n)$ in different ways.
On the one hand, we can determine the Dirichlet inverse of $\alpha(n) = (-1)^{n-1}$. Since $\alpha$ is multiplicative, that is relatively easy, one finds
$$b(n) = \begin{cases} \mu(n) &, n \equiv 1 \pmod{2} \\ 2^{v_2(n)-1}\mu\left(\dfrac{n}{2^{v_2(n)}}\right)&, n \equiv 0 \pmod{2}. \end{cases}\tag{1}$$
On the other hand, from the product representation
$$\begin{align}
\eta(s) &= \left(1 - \frac{1}{2^{s-1}}\right)\prod_p \left(1 - \frac{1}{p^s}\right)^{-1}\\
&= \frac{2^s-2}{2^s-1} \prod_{p > 2} \left(1 - \frac{1}{p^s}\right)^{-1}
\end{align}$$
by expanding
$$\frac{2^s-1}{2^s-2} = 1 + \frac{1}{2^s-2} = 1 + \sum_{\nu = 1}^\infty \frac{2^{\nu-1}}{2^{\nu s}},$$
we obtain the product representation
$$\frac{1}{\eta(s)} = \left(1 + \sum_{\nu=1}^\infty \frac{2^{\nu-1}}{2^{\nu s}}\right)\prod_{p > 2} \left(1 - \frac{1}{p^s}\right),$$
which is easily seen to lead to $(1)$.
A: $$\frac{1}{\eta(s)}=\frac1{(1-2^{1-s})\cdot\zeta(s)}=\frac1{1-2^{1-s}}\cdot\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{s}}=\sum_{n=1}^{\infty}\frac{b(n)}{n^{s}}\iff b(n)=\frac{\mu(n)}{1-2^{1-s}}$$
A: You are correct that $b(n)$ can be computed recursively as the Dirichlet inverse of $a(n)$. But there's another way: simply expand $(1-2^{1-s})^{-1}$ and multiply that against $\zeta(s)^{-1}$:
$$\frac{1}{\eta(s)}=\cfrac{1}{\displaystyle 1-\frac{2}{2^s}}\frac{1}{\zeta(s)}=\left(1+\frac{2}{2^s}+\frac{4}{4^s}+\frac{8}{8^s}+\cdots\right)\sum_{n=1}^\infty\frac{\mu(n)}{n^s}=\sum_{n=1}^\infty\frac{b(n)}{n^s} $$
Therefore, for $n$ odd, $b(n)=\mu(n)$, and for $n$ even,
$$\begin{array}{ll} b(n) & =\sum_{\substack{d\mid n \\ d=2^r}}d\mu(n/d) \\ & =\sum_{r=0}^{v}2^r\mu(n/2^r) \\ & =2^{v_2(n)}\mu(n/2^{v_2(n)})+2^{v_2(n)-1}\mu(n/2^{v_2(n)-1}) \\[3pt] & =2^{v_2(n)-1}\mu(n/2^{v_2(n)}).\end{array}$$
