The Möbius Inversion for an infinite series(assuming absolute convergence) is given by, if \begin{equation} b(n) = \sum_{k=1}^\infty a(kn) \iff a(n) = \sum_{k=1}^\infty \mu(k)\, b(kn) \end{equation} The proof is given as \begin{equation} \sum_{k=1}^\infty \mu(k)\, \sum_{d=1}^\infty a(dkn) = \sum_{N=1}^\infty a(Nn)\, \sum_{k|N}\mu(k) = a(n) \end{equation}

My question is how we use above if we have a series of the type \begin{equation} b(n) = \sum_{k=1}^\infty a(kn)\,x^k \end{equation} Can we say that \begin{equation} b(n) = \sum_{k=1}^\infty a(kn)\,x^k \iff a(n)x = \sum_{k=1}^\infty \mu(k)\, b(kn) \qquad \qquad ? \end{equation} I think I am wrong but have no idea how to get Möbius Inversion of the above type series.? Thanks for your help.

  • $\begingroup$ No. But $\sum_{n=1}^\infty \mu(n) \sum_{k=1}^\infty a(kn) x^{kn} = ?$ $\endgroup$ – reuns Aug 13 '17 at 11:48
  • $\begingroup$ Also with $b(n) = \sum_{k=1}^\infty a(kn) x^k$ then $a(n) = \sum_{k=1}^\infty y_k b(nk)$ where $y_k$ is the Dirichlet inverse of $x^k$ ie. $\sum_{d | k} y_d x^{k/d} = 1_{k = 1}$ $\endgroup$ – reuns Aug 13 '17 at 11:51

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