Determine the radius of convergence 
of the power series:
![\[ \sum_{n=1}^{\infty} \left( \frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot \cdots \cdot (2n)} \right)^3 z^n. \]](http://s13997.pcdn.co/wp-content/ql-cache/quicklatex.com-ddb9e0ddcfd72cd627faeacf8e34772b_l3.png "Rendered by QuickLaTeX.com")
Test for convergence at the boundary points if is finite.
We have
![\[ a_n = \left( \frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot \cdots \cdot (2n)} \right)^3 z^n. \]](http://s13997.pcdn.co/wp-content/ql-cache/quicklatex.com-091fb913c9f9c037fcf4eaaeacd78821_l3.png "Rendered by QuickLaTeX.com")
Then applying the ratio test we have
![\begin{align*} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \left( \frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot (2n+1)}{2 \cdot 4 \cdot 6 \cdot \cdots \cdot (2n+2)} \right)^3 \cdot z^{n+1} \cdot \left( \frac{2 \cdot 4 \cdot 6 \cdot \cdots \cdot (2n)}{1 \cdot 3 \cdot 5 \cdot \cdots \cdot (2n-1)} \right)^3 \left( \frac{1}{z^n} \right) \\[9pt] &= \lim_{n \to \infty} \left( \frac{(2n+1)^3}{(2n+2)^3} \right) z \\ &= z. \end{align*}](http://s13997.pcdn.co/wp-content/ql-cache/quicklatex.com-9f6f40c80f0139c388ca77ac28989ada_l3.png "Rendered by QuickLaTeX.com")
Hence, the radius of convergence is 1, and the series converges for 
.