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Determine the radius of convergence of ∑ (-1)n22nz2n / 2n

Determine the radius of convergence r of the power series:

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

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


This is an alternating series, and we have

    \[ \left| \frac{(-1)^n 2^{2n} z^{2n}}{2n} \right| = \frac{|2z|^{2n}}{2n}. \]

If |2z| > 1 then the terms are not going to 0 so the series is divergent. If |2z| < 1 then it is monotonically decreasing, so we know (by the Leibniz rule) that the series converges if and only if

    \[ \lim_{n \to \infty} \frac{|2z|^{2n}}{2n} = 0 \quad \iff \quad |z| \leq \frac{1}{2}. \]

Therefore, the radius of convergence is r = \frac{1}{2}, and the series converges at every boundary point |z| = \frac{1}{2}.

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