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Determine the radius of convergence of ∑ ((n!)2 / (2n!)) zn

Determine the radius of convergence r of the power series:

    \[ \sum_{n=1}^{\infty} \frac{(n!)^2}{(2n!)} z^n. \]

Test for convergence at the boundary points if r is finite.


Let a_n = \frac{(n!)^2 z^n}{(2n)!}. Then we have

    \begin{align*}  \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{(n+1)!^2 z^{n+1}}{(2n+2)!} \right) \left( \frac{(2n)!}{(n!)^2 z^n} \right) \\[9pt]  &= \lim_{n \to \infty} \frac{ (n+1)^2}{(2n+1)(2n+2)} \right) z\\  &= \frac{z}{4}. \end{align*}

Thus, the radius of convergence is r = 4. This diverges on the boundary points since if |z| = 4 then the terms do not go to 0.

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