I'm trying to invert the following power series \begin{equation} \omega(x) = \frac{6}{5}\frac{1}{x^5} - \sum_{k = 1}^\infty \frac{6}{4k - 5}A_ks^kx^{4k - 5} \end{equation} so that I get $x = x(\omega)$. In this series, $s$ is an unknown constant and $A_k$ is a series of unknown constants for $k = 1,2,3,...$. Mathematica will have no problem defining this series and invert it when I specified what $A_k$ is. For example I can let $A_k = k, \forall k$ then
Series[(6/5) x^(-5) - Sum[(6 /(4 n - 5)) A[n] s^n x^(4 n - 5), {n, 1,Infinity}], {x, 0, 20}]
will produce the following desirable output
\begin{equation} \frac{6}{5 x^5}+\frac{6s}{x}-2s^2 x^3-\frac{6s^3 x^7}{7}-\frac{6s^4x^{11}}{11}-\frac{2s^5 x^{15}}{5}-\frac{6s^6 x^{19}}{19}+O\left(x^{21}\right) \end{equation}
Then I can use InverseSeries[%] to produces the inverse. However what I really want is to not specify $A_k$ and get a series of the form \begin{equation} \frac{6}{5 x^5}+\frac{6 A[1] s}{x}-2 A[2]s^2 x^3-\frac{6 A[3]s^3 x^7}{7}-\frac{6 A[4]s^4x^{11}}{11}-\frac{2 A[5]s^5 x^{15}}{5}-\frac{6 A[6]s^6 x^{19}}{19}+O\left(x^{21}\right) \end{equation} then invert it in term of these unknown $A[1],A[2],A[3],...$. However when I typed the same code as above without specifying $A_k$, I get \begin{equation} (\frac{6}{5 x^5} + O(x^{21})) - \sum_{n=1}^\infty \frac{6 A[n] s^n x^{4n - 5}}{4n - 5} \end{equation} and InverseSeries[%] will not do anything for me. What is the way to achieve what I'm trying to do?
Series[(6/5) x^(-5) - Sum[(6/(4 n - 5)) A[n] s^n x^(4 n - 5), {n, 1, 20}], {x, 0, 20}]
? $\endgroup$Infinity
inSum
with the order of the series sought. $\endgroup$