Question

The radius of convergence of the power series \[ \sum_{n=1}^{\infty} \frac{(x + \frac{1}{4})^n}{(-2)^n n^2} \] about \( x = -\frac{1}{4} \) is equal to ............ (rounded off to two decimal places).

Hint:

The ratio test is a powerful method for determining the radius of convergence of a power series. The limit of the ratio of consecutive terms gives \( \frac{1}{R} \), where \( R \) is the radius of convergence.

Answer 1

Step 1: Use the ratio test for the radius of convergence.
We apply the ratio test to find the radius of convergence \( R \). The general form of the power series is: \[ \sum_{n=1}^{\infty} a_n (x - c)^n, \] where \( a_n \) are the coefficients and \( c = -\frac{1}{4} \) is the center of the series. The given series is: \[ a_n = \frac{1}{(-2)^n n^2}. \] The ratio test for the radius of convergence \( R \) involves: \[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \frac{1}{R}. \] Step 2: Compute the ratio.
We compute the ratio of consecutive terms \( a_{n+1} \) and \( a_n \): \[ \frac{a_{n+1}}{a_n} = \frac{\frac{1}{(-2)^{n+1} (n+1)^2}}{\frac{1}{(-2)^n n^2}} = \frac{n^2}{(-2)(n+1)^2}. \] Simplifying, we get: \[ \frac{a_{n+1}}{a_n} = \frac{n^2}{(-2)(n+1)^2}. \] Now, take the limit as \( n \to \infty \): \[ \lim_{n \to \infty} \left| \frac{n^2}{(-2)(n+1)^2} \right| = \lim_{n \to \infty} \frac{n^2}{2(n+1)^2} = \frac{1}{2}. \] Thus, the radius of convergence \( R \) is: \[ R = 2. \] Final Answer: \[ \boxed{R = 2}. \]