Home » Blog » Determine the radius of convergence of ∑ ((1*3*5*…*(2n-1)) / (2*4*6*…*(2n)))3 z

Determine the radius of convergence of ∑ ((1*3*5*…*(2n-1)) / (2*4*6*…*(2n)))3 z

Determine the radius of convergence r of the power series:

    \[ \sum_{n=1}^{\infty} \left( \frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot \cdots \cdot (2n)} \right)^3 z^n. \]

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


We have

    \[ a_n = \left( \frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot \cdots \cdot (2n)} \right)^3 z^n. \]

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*}

Hence, the radius of convergence is 1, and the series converges for |z| < 1.

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