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For each $m \in \mathbb{N}$, define the power series as $\sum_{n} \binom{mn}{n} z^n$, find its radius of convergence. I tried with ratio test but I got really messy algebra.

My attempt so far: Let $R$ denote the radius of convergence, then $R = (\lim _{n\to \infty} \frac{(mn+m)!(mn-n)!}{n((m-1)(n+1))!(mn)!})^{-1}$ then should I try to expand this and cancel as much as I can?

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    $\begingroup$ Can you write down your ratio test attempt? It doesn't look that messy to me. EDIT: I wrote down the ratio test and it seems to provide conclusive results; just collect the coefficients of the highest powers of m to simplify your expression. $\endgroup$ – Alexander Geldhof Jun 16 at 12:48
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    $\begingroup$ This is not clear at all: is $\;m\;$ a constant number? If not, how does it play a role in the power series $\;\sum\limits_{n=1}^\infty\binom{mn}n z^n\;$ ? Perhaps a double series...? $\endgroup$ – DonAntonio Jun 16 at 12:48
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    $\begingroup$ @DonAntonio As is stated at the start of the problem, you're supposed to solve the problem for each $m \in \mathbb{N}$. $\endgroup$ – Alexander Geldhof Jun 16 at 12:51
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    $\begingroup$ @DonAntonio I am indeed not the OP, but the wording is clear and non ambiguous. It says 'for each $m \in \mathbb{N}$, define the power series... Ergo, we're defining a class of power series uncountable in number, not one power series. $\endgroup$ – Alexander Geldhof Jun 16 at 12:53
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    $\begingroup$ @AlexanderGeldhof Well, perhaps...yet the OP hasn't yet addressed the doubt. $\endgroup$ – DonAntonio Jun 16 at 12:57
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I'd use the Cauchy-Hadamard theorem (basically the root test) instead, which has the advantage of having fewer factorials to manipulate.

So the radius of convergence $R$ is given by $$ \frac1R=\limsup_{n\to\infty}\binom{mn}{n}^{1/n}=\limsup_{n\to\infty}\frac{(mn)!^{1/n}}{n!^{1/n}[(m-1)n]!^{1/n}} $$ but Stirling's formula $n!=\sqrt{2\pi n}(n/e)^{n}(1+O(n^{-1}))$ gives $(mn)!^{1/n}=(2\pi mn)^{1/(2n)} (mn/e)^m(1+o(n^{-1}))$. Hence $$ \frac1R=\limsup_{n\to\infty}\frac{(2\pi mn)^{1/(2n)} (mn/e)^m}{(2\pi n)^{1/(2n)} (n/e)(2\pi (m-1)n)^{1/(2n)} ((m-1)n/e)^{m-1}}=\frac{m^m}{(m-1)^{m-1}}. $$

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Using the Ratio test, and assuming $m > 1$:

$\begin{align*} \frac{{mn + m \choose n+1}}{{mn \choose n}} &= \frac{(mn + m)! (n)!((m-1)n)!}{(mn)!(n+1)!((m-1)(n+1))!} \\ &= \frac{(mn+m)!}{(mn)!} \frac{n!}{(n+1)!}\frac{(m n - n)!}{(mn - n + m - 1)!} \\ &= (mn + 1)\ldots(mn + m) \frac{1}{n+1} \frac{1}{(mn - n + 1)\ldots(mn - n + m - 1)}\\ &= \frac{m(n + 1)}{n + 1} \frac{m\cdot n + 1}{(m - 1)\cdot n + 1} \ldots \frac{m\cdot n + m - 1}{(m - 1)\cdot n + m - 1} \end{align*}$

and taking the limits of the fractions separately, we arrive at the same answer as @user10354138 did.

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