Question

The radius of convergence of the power series \(\sum_{n=1}^\infty\left(\frac{2^n}{n}+\frac{3^n}{n^2}\right)z^n\) is ____.

• A. \(e\)
• B. \(1\)
• C. \(\dfrac{1}{2}\)
• D. \(\dfrac{1}{3}\)

Hint:

Compare growth rates of \(2^n/n\) and \(3^n/n^2\); the faster-growing term controls the radius.

Answer 1

The Correct Option is D

Step 1: Write the coefficient of \(z^n\) as \(a_n=\dfrac{2^n}{n}+\dfrac{3^n}{n^2}\). By the Cauchy-Hadamard formula, the radius of convergence is \(R=\dfrac{1}{\limsup_{n\to\infty}|a_n|^{1/n}}\).

Step 2: Compare the two terms as \(n\to\infty\). Since \(\left(\dfrac{3}{2}\right)^n\to\infty\), the ratio \(\dfrac{3^n/n^2}{2^n/n}=\dfrac{1}{n}\left(\dfrac{3}{2}\right)^n\to\infty\), so \(\dfrac{3^n}{n^2}\) dominates \(\dfrac{2^n}{n}\), giving \(a_n\sim\dfrac{3^n}{n^2}\) for large \(n\).

Step 3: Then \(|a_n|^{1/n}\sim\left(\dfrac{3^n}{n^2}\right)^{1/n}=3\cdot n^{-2/n}\). Since \(n^{1/n}\to1\) as \(n\to\infty\), \(n^{-2/n}\to1\), so \(\limsup_{n\to\infty}|a_n|^{1/n}=3\).

Step 4: Therefore \(R=\dfrac{1}{3}\).

\[\boxed{R=\dfrac{1}{3}}\]