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If you're worried about convergence... You can rewrite your equation as $g(z) = {1 \over f(1/z)} = 1/n$ for $z$ near ${1 \over n}$. $g(z)$ is thus $z^d$ divided by a polynomial with a nonzero constant term. Here $d$ is the degree of $f(z)$. Hence $g(z) = h(z)^d$ for some $h(z)$ with $h(0) = 0$ and $h'(0) \neq 0$. So you're solving $h(z) = n^{-{1 \over d}}$ which has a convergent power series solution in $n^{-{1 \over d}}$ by inverting $h(z)$.

I should mention, this gives $g(z)$, but then $f(z) = {1 \over g(1/z)}$ which gives that $f(z)$ has an expansion starting with $cz^{1 \over d}$ and then the exponents decrease by multiples of ${1 \over d}$.

If you're worried about convergence... You can rewrite your equation as $g(z) = {1 \over f(1/z)} = 1/n$ for $z$ near ${1 \over n}$. $g(z)$ is thus $z^d$ divided by a polynomial with a nonzero constant term. Here $d$ is the degree of $f(z)$. Hence $g(z) = h(z)^d$ for some $h(z)$ with $h(0) = 0$ and $h'(0) \neq 0$. So you're solving $h(z) = n^{-{1 \over d}}$ which has a convergent power series solution in $n^{-{1 \over d}}$ by inverting $h(z)$. Then the series you originally wanted will be the reciprocal of this, so will start with $cn^{1 \over d}$ and then the exponents decrease by multiples of ${1 \over d}$.

If you're worried about convergence... You can rewrite your equation as $g(z) = {1 \over f(1/z)} = 1/n$ for $z$ near ${1 \over n}$. $g(z)$ is thus $z^d$ divided by a polynomial with a nonzero constant term. Here $d$ is the degree of $f(z)$. Hence $g(z) = h(z)^d$ for some $h(z)$ with $h(0) = 0$ and $h'(0) \neq 0$. So you're solving $h(z) = n^{-{1 \over d}}$ which has a convergent power series solution in $n^{-{1 \over d}}$ by inverting $h(z)$.

If you're worried about convergence... You can rewrite your equation as $g(z) = {1 \over f(1/z)} = 1/n$ for $z$ near ${1 \over n}$. $g(z)$ is thus $z^d$ divided by a polynomial with a nonzero constant term. Here $d$ is the degree of $f(z)$. Hence $g(z) = h(z)^d$ for some $h(z)$ with $h(0) = 0$ and $h'(0) \neq 0$. So you're solving $h(z) = n^{-{1 \over d}}$ which has a convergent power series solution in $n^{-{1 \over d}}$ by inverting $h(z)$.

I should mention, this gives $g(z)$, but then $f(z) = {1 \over g(1/z)}$ which gives that $f(z)$ has an expansion starting with $cz^{1 \over d}$ and then the exponents decrease by multiples of ${1 \over d}$.

If you're worried about convergence... You can rewrite your equation as $g(z) = {1 \over f(1/z)} = 1/n$ for $z$ near ${1 \over n}$. $g(z)$ is thus $z^d$ divided by a polynomial with a nonzero constant term. Here $d$ is the degree of $g(z)$. Hence $g(z) = h(z)^d$ for some $h(z)$ with $h(0) = 0$ and $h'(0) \neq 0$. So you're solving $h(z) = n^{-{1 \over d}}$ which has a convergent power series solution in $n^{-{1 \over d}}$ by inverting $h(z)$.

If you're worried about convergence... You can rewrite your equation as $g(z) = {1 \over f(1/z)} = 1/n$ for $z$ near ${1 \over n}$. $g(z)$ is thus $z^d$ divided by a polynomial with a nonzero constant term. Here $d$ is the degree of $f(z)$. Hence $g(z) = h(z)^d$ for some $h(z)$ with $h(0) = 0$ and $h'(0) \neq 0$. So you're solving $h(z) = n^{-{1 \over d}}$ which has a convergent power series solution in $n^{-{1 \over d}}$ by inverting $h(z)$.