Question

The radius of convergence of the power series \( \displaystyle \sum_{n=1}^{\infty} \left( \dfrac{n+2}{n} \right)^{n^2} x^n \) is
 

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

Hint:

For power series, the root test is the fastest way to find the radius of convergence: \( R = 1 / \limsup |a_n|^{1/n}. \)

Answer 1

The Correct Option is D

Step 1: Apply root test.
Let \( a_n = \left( \dfrac{n+2}{n} \right)^{n^2}. \) Then radius of convergence \( R = \dfrac{1}{\limsup_{n \to \infty} |a_n|^{1/n}}. \)

Step 2: Simplify \( |a_n|^{1/n} \).
\( |a_n|^{1/n} = \left( \dfrac{n+2}{n} \right)^n = \left( 1 + \dfrac{2}{n} \right)^n. \)

Step 3: Take the limit.
\( \lim_{n \to \infty} \left( 1 + \dfrac{2}{n} \right)^n = e^2. \)

Step 4: Find radius of convergence.
\( R = \dfrac{1}{e^2}. \)

Final Answer: \( R = \dfrac{1}{e^2}. \)