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Let $\mu$ be Mobius function, defined by $\mu(n)=\begin{cases} (-1)^{\omega(n)}, \text{ if } n \text { is square free}\\ 0, \text{ otherwise} \end{cases}$, where $\omega(n)$ is the number of prime factors of $n$. Also, consider the $k$-th divisor function $\tau_k(n)$, defined as the number of representations of $n$ as product of $k$ integers, $\tau_k(n)=\sum_{d_1d_2\dots d_k=n}1$.

Consider the generating Dirichlet series for pointwise product $\mu(n)\tau_k(n)$, $$ D(s)=\sum_{n=1}^\infty \frac{\mu(n)\tau_k(n)}{n^s}. $$ I am interested in expressing this series in terms of Riemann zeta function $\zeta(s)=\sum_{n=1}^\infty\frac{1}{n^s}$. Using the Euler product for $D(s)$ I get \begin{align*} D(s)&=\prod_{p} \left( 1+\sum_{i=1}^\infty \frac{\mu(p^i)\tau_k(p^i)}{p^{is}} \right)\\ &=\prod_{p} \left( 1-\frac{k}{p^s} \right ), \end{align*} but then I don't know how to connect the expression I got with $\zeta(s)$.

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Twists by completely multiplicative functions work better $$\zeta(s)^k= \sum_{n=1}^\infty \tau_k(n) n^{-s}, \qquad \frac{\zeta(2s)}{\zeta(s)} = \sum_{n=1}^\infty s(n) n^{-s}$$ where $s(n)= \prod_{p^m \| n} (-1)^m$ is completely multiplicative so $$\frac{\zeta(2s)^k}{\zeta(s)^k} = \sum_{n=1}^\infty s(n) \tau_k(n)n^{-s}$$


Now $$ T_k(s)=\sum_{n=1}^\infty \mu(n) \tau_k(n)n^{-s}= \prod_p (1-k p^{-s})= \frac{\zeta(2s)^k}{\zeta(s)^k}\prod_p (1-k p^{-s})(1+p^{-s})^k \\= \frac{\zeta(2s)^k}{\zeta(s)^k} U_k(s)$$ where $$U_k(s) = \prod_p(1+\sum_{l=2}^{k+1} p^{-sl} ({k \choose l}-k{k-1 \choose l})$$ is analytic for $\Re(s) > 1/2$.


As $$ \log T_k(s) = -\sum_{m=1}^\infty \log\zeta(sm)(\sum_{d | m} \mu(d)\frac{k^{m/ d}}{m/d})$$ you can find the analytic continuation of $T_k$ and its singularities and show it is has natural boundary at $\Re(s) = 0$.

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