Determine the radius of convergence 
of the power series:
![\[ \sum_{n=1}^{\infty} \frac{(-1)^n 2^{2n} z^{2n}}{2n}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-e2ac4a3906fb04a86cfd553e27ac720e_l3.png "Rendered by QuickLaTeX.com")
Test for convergence at the boundary points if is finite.
This is an alternating series, and we have
![\[ \left| \frac{(-1)^n 2^{2n} z^{2n}}{2n} \right| = \frac{|2z|^{2n}}{2n}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-c23b57f8a7d3546f5ee8f298526c3ca2_l3.png "Rendered by QuickLaTeX.com")
If 
then the terms are not going to 0 so the series is divergent. If 
then it is monotonically decreasing, so we know (by the Leibniz rule) that the series converges if and only if
![\[ \lim_{n \to \infty} \frac{|2z|^{2n}}{2n} = 0 \quad \iff \quad |z| \leq \frac{1}{2}. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-0e905cc1439937e5197fbc5425fe3058_l3.png "Rendered by QuickLaTeX.com")
Therefore, the radius of convergence is 
, and the series converges at every boundary point 
.
The series diverges at
. Refer to theorem 10.19.
Also For z = -i/2