Discussion
The radius of convergence of the series
\[ 1 - 4z^{2} + 16z^{4} - 64z^{6} + 256z^{8} - \cdots \]
is :
Answer: D
The radius of convergence is \(\frac{1}{2}\).
The series is geometric in \(-4z^{2}\):
\[ \begin{aligned} &1 - 4z^{2} + 16z^{4} - \cdots \\ &= \sum_{n=0}^{\infty} (-4z^{2})^{n} = \frac{1}{1 + 4z^{2}} \end{aligned} \]
It converges exactly when \(|4z^{2}| \lt 1\), that is, \(|z| \lt 1/2\). Equivalently, the function \(1/(1 + 4z^{2})\) has poles at \(z = \pm i/2\), and the radius of convergence about 0 is the distance to the nearest singularity, 1/2.
The value 2 is the reciprocal error (treating 4 as the radius squared rather than its inverse), 1 ignores the coefficients altogether, and \(\infty\) would require an entire function.
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