Discussion
If \(R\) denotes the radius of convergence of the power series
\[ \begin{aligned} \sum_{n=1}^{\infty} &\left(\frac{2 \cdot 4 \cdot 6 \cdots 2n}{2 \cdot 5 \cdot 8 \cdots (3n-1)}\right)^{2} \\ &\times x^{n}, \end{aligned} \]
then which one of the following statements is true?
Answer: B
The radius of convergence is \(R = \frac{9}{4}\), so \(4R = 9\).
Use the ratio test on the coefficients \(a_{n}\), the squared fractions. Going from \(n\) to \(n+1\) adds one factor \(2n+2\) on top and \(3n+2\) underneath:
\[ \begin{aligned} \frac{a_{n+1}}{a_{n}} &= \left(\frac{2n+2}{3n+2}\right)^{2} \\ &\to \left(\frac{2}{3}\right)^{2} = \frac{4}{9} \end{aligned} \]
so \(R = \lim \frac{a_{n}}{a_{n+1}} = \frac{9}{4}\). The option \(9R = 4\) gives \(R = \frac{4}{9}\), the reciprocal, which is the limit of the ratio itself rather than of its inverse. \(R = 1\) and \(R = \frac{1}{2}\) come from forgetting the square or misreading the factors.
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