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If $|z| > 2$ it's easy to see that this series is divergent. Can you see why? If $|z| < 2$ it's easy to see that this series is convergent. Can you see why? So the radius of convergence is $2$.

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An important notion is the algebraic order of a meromorphic function at a point. If $f$ is meromorphic on the connected open set $U \subset \mathbb{C}$, not vanishing identically, and $z_0\in U$, then the algebraic order of $f$ at $z_0$, denoted by $\mathscr{o}_{z_0}(f)$ or $\mathscr{o}(f; z_0)$ (or similar) is defined to be the (unique) $k \in \mathbb{Z}$ ...

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The radius of convergence of a power series $\sum_{n=0}^\infty a_n z^n$ with complex coefficients is \begin{equation*} R \,=\, \liminf_{n\to\infty}\, |a_n|^{-1/n}~. \end{equation*} $($If $a_n=0$, then $|a_n|^{-1/n}=+\infty$.$)\,$ In your case the radius of convergence is \begin{equation*} R \,=\, \liminf_{n\to\infty}\, \left|\frac{\log(n)}{n\cdot ...

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Since $f(z)=\frac{z^2}{(z-a_1)(z-a_2)(z-a_3)(z-a_4)}$, $$(z-a_1)f(z)=\frac{z^2}{(z-a_2)(z-a_3)(z-a_4)}.$$ But the right-hand side is well-defined and continuous at $z=a_1$, so we get that $$\lim_{z\to a_1}(z-a_1)f(z)=\frac{a_1^2}{(a_1-a_2)(a_1-a_3)(a_1-a_4)}.$$

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