Question

Let \( a_n = \dfrac{(1 + (-1)^{n})}{2^n} + \dfrac{(1 + (-1)^{n-1})}{3^n}. \) Then the radius of convergence of the power series \( \sum_{n=1}^{\infty} a_n x^n \text{ about } x = 0 \text{ is ..............} \)

Hint:

The ratio test is often used to determine the radius of convergence for power series.

Answer 1

Correct Answer: 2

To determine the radius of convergence of the series \(\sum_{n=1}^{\infty} a_n x^n\), we start by analyzing the expression for \(a_n\):

\(a_n = \dfrac{(1 + (-1)^n)}{2^n} + \dfrac{(1 + (-1)^{n-1})}{3^n}\).
Note that for even \(n\), \(1 + (-1)^n = 2\) and for odd \(n\), \(1 + (-1)^n = 0\). Similarly, for even \(n\), \(1 + (-1)^{n-1} = 0\) and for odd \(n\), \(1 + (-1)^{n-1} = 2\).

Thus, for even \(n\), \(a_n = \dfrac{2}{2^n}\) and for odd \(n\), \(a_n = \dfrac{2}{3^n}\).
This can be simplified to:

• When \(n\) is even: \(a_n = \dfrac{1}{2^{n-1}}\).
• When \(n\) is odd: \(a_n = \dfrac{2}{3^n}\).

To find the radius of convergence \(R\), use the formula:

\(\dfrac{1}{R} = \limsup_{n \to \infty} |a_n|^{1/n}\).

Calculate separately for even and odd \(n\):

• Even \(n\): \(|a_n|^{1/n} = \left(\dfrac{1}{2^{n-1}}\right)^{1/n} = \dfrac{1}{2^{(n-1)/n}}\), and as \(n \to \infty\), this approaches \(\dfrac{1}{2}\).
• Odd \(n\): \(|a_n|^{1/n} = \left(\dfrac{2}{3^n}\right)^{1/n} = \dfrac{2^{1/n}}{3}\), and as \(n \to \infty\), this approaches \(\dfrac{1}{3}\).

The larger quantity dominates, so \(\limsup_{n \to \infty} |a_n|^{1/n} = \dfrac{1}{2}\).

Therefore, \(\dfrac{1}{R} = \dfrac{1}{2}\) or \(R = 2\).

Consequently, the radius of convergence is 2.