The radius of convergence of the power series \( \displaystyle \sum_{n=1}^{\infty} \left( \dfrac{n+2}{n} \right)^{n^2} x^n \) is
The radius of convergence of the power series \( \displaystyle \sum_{n=1}^{\infty} \left( \dfrac{n+2}{n} \right)^{n^2} x^n \) is
Show Hint
- \( e^2 \)
- \( \dfrac{1}{\sqrt{e}} \)
- \( \dfrac{1}{e} \)
- \( \dfrac{1}{e^2} \)
The Correct Option is D
Solution and Explanation
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: (D) \( R = \dfrac{1}{e^2}. \)
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