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Is there any way to translate $\sum_n \frac{x^n}{a_n}$ into a generating function of type $A(x)$ or into any combination involving $A(x)$? This question comes from a treatment I'm giving to equation $$-\frac{A(x)-a_0}{x}+\frac{2}{1-x}=\sum_n\frac{x^n}{a_n}\qquad \qquad (1)$$in an attempt to find a simple generating function whose expansion might yield the sequence for the following non-linear recurrence equation:$$a_{n+1}=2-\frac{1}{a_n},$$given the initial value $a_0$.

Would there be any relation to harmonic series?

Would Cauchy's product rule help?

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    $\begingroup$ There is no need to use generating function. By writing $a_n$ as $\frac{p_n}{q_n}$, you can transform your problem to one involving a linear recurrence relation instead! $$\begin{bmatrix}p_{n+1}\\q_{n+1}\end{bmatrix} = \begin{bmatrix}2&-1\\1&0\end{bmatrix}\begin{bmatrix}p_{n}\\q_{n}\end{bmatrix}$$ $\endgroup$ – achille hui Jun 4 '14 at 17:00
  • $\begingroup$ @achille hui: I do appreciate your answer but my central question was, if $\sum_n a_n x^n$ is associated to a generating function $A(x)$, what is $\sum_n 1/a_n x^n$ associated to? $\endgroup$ – Tavasanis Jun 5 '14 at 7:22
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There are a number of ways of seeing that you're unlikely to get any 'clean' relation: for instance, $\sum_nnx^n$ is a rational function of $x$, $\frac{x}{(1-x)^2}$, but $\sum_n\frac{x^n}{n}$ is $\ln(1-x)$. Worse, we have $\sum_n\frac{x^n}{n!}=e^x$, but $F(x) =\sum_nn!x^n$ is a nowhere-convergent series (formally it solves the differential equation $x^2\frac{dF}{dx} = (1-x)F(x)-1$, but this equation is ill-behaved at $x=0$).

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There is no simple relation between $\sum_{n \ge 0} a_n x^n$ and $\sum_{n \ge 0} \frac{x^n}{a_n}$. Just as there is no simple relation between $\sum_{n \ge 0} a_n x^n$, $\sum_{n \ge 0} b_n x^n$ and $\sum_{n \ge 0} a_n b_n x^n$.

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