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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?

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    $\begingroup$ Series[(6/5) x^(-5) - Sum[(6/(4 n - 5)) A[n] s^n x^(4 n - 5), {n, 1, 20}], {x, 0, 20}]? $\endgroup$ – Michael E2 Apr 25 '17 at 11:00
  • $\begingroup$ Something wrong with the code I used? $\endgroup$ – user113988 Apr 25 '17 at 11:14
  • $\begingroup$ I replaced Infinity in Sum with the order of the series sought. $\endgroup$ – Michael E2 Apr 25 '17 at 11:16
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The problem means extraction of coefficients from a given power series.

Redefine your function $\omega (x)$ with a finite sum

\[Omega][x_, m_] := 
 6/(5 x^5) - Sum[(6 s^n x^(-5 + 4 n) A[n])/(-5 + 4 n), {n, 1, m}]

Then, for any finite $m$, you can peform the extraction procedure using Coeffient[] as follows

With[{m = 20}, Table[{k, Coefficient[\[Omega][x, m], x, k]}, {k, -5, 20}]]

(* Out[104]= {{-5, 6/5}, {-4, 0}, {-3, 0}, {-2, 0}, {-1, 6 s A[1]}, {0, 0}, {1, 0}, {2, 0}, {3, -2 s^2 A[2]}, {4, 0}, {5, 0}, {6, 0}, {7, -(6/7) s^3 A[3]}, {8, 
  0}, {9, 0}, {10, 0}, {11, -(6/11) s^4 A[4]}, {12, 0}, {13, 0}, {14, 
  0}, {15, -(2/5) s^5 A[5]}, {16, 0}, {17, 0}, {18, 
  0}, {19, -(6/19) s^6 A[6]}, {20, 0}} *)

A more compact form which gives directly the coefficients $A(k)$ is

With[{m = 20}, 
 Cases[Table[{k, Coefficient[\[Omega][x, m], x, k]}, {k, -5, 
    20}], {_, _A _}]]

(* Out[126]= {{-1, 
  6 s A[1]}, {3, -2 s^2 A[2]}, {7, -(6/7) s^3 A[3]}, {11, -(6/11)
     s^4 A[4]}, {15, -(2/5) s^5 A[5]}, {19, -(6/19) s^6 A[6]}} *)
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