Question

The radius of convergence of the series \[ \sum_{n \geq 0} 3^{n+1} z^{2n}, \quad z \in \mathbb{C} \] is __________ (round off to TWO decimal places).

Hint:

For a power series of the form \( \sum a_n z^{2n} \), use the Ratio Test to find the radius of convergence. The radius of convergence \( R \) is given by \( R = \frac{1}{\lim \left| \frac{a_{n+1}}{a_n} \right|} \).

Answer 1

Correct Answer: 0.55

We are given the power series: \[ \sum_{n \geq 0} 3^{n+1} z^{2n} \] This is a series in the form of \[ \sum_{n \geq 0} a_n z^{2n}, \quad \text{where} \quad a_n = 3^{n+1} \] To find the radius of convergence, we use the Root Test or the Ratio Test. Here, we will use the Ratio Test for the series. The Ratio Test for the series \[ \sum a_n z^{2n} \] gives the radius of convergence \( R \) as: \[ \frac{1}{R} = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \] Now, calculate the ratio \( \frac{a_{n+1}}{a_n} \): \[ \frac{a_{n+1}}{a_n} = \frac{3^{(n+2)}}{3^{n+1}} = 3 \] Thus, \[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = 3 \] So, the radius of convergence \( R \) is: \[ R = \frac{1}{3} \] Therefore, the radius of convergence of the series is approximately \[ R \approx 0.33 \]