Question

Let \( \{a_n\}_{n=0}^{\infty} \) and \( \{b_n\}_{n=0}^{\infty} \) be sequences of positive real numbers such that \( n a_n<b_n<n^2 a_n \), for all \( n \geq 2 \). If the radius of convergence of the power series \[ \sum_{n=0}^{\infty} a_n x^n \] is 4, then the power series \[ \sum_{n=0}^{\infty} b_n x^n \] is

• A. converges for all \( x \) with \( |x|<2 \)
• B. converges for all \( x \) with \( |x|>2 \)
• C. does not converge for any \( x \) with \( |x|>2 \)
• D. does not converge for any \( x \) with \( |x|<2 \)

Hint:

When comparing two power series, the radius of convergence depends on the growth rate of the coefficients. Faster-growing coefficients lead to a smaller radius of convergence.

Answer 1

The Correct Option is A

Step 1: Understanding the conditions.
We are given that the power series \( \sum_{n=0}^{\infty} a_n x^n \) has a radius of convergence of 4, meaning it converges for \( |x|<4 \). We are also told that \( n a_n<b_n<n^2 a_n \) for all \( n \geq 2 \).

Step 2: Comparing the series.
Since \( b_n \) grows faster than \( a_n \), the radius of convergence of the series \( \sum_{n=0}^{\infty} b_n x^n \) will be smaller than that of the series for \( a_n x^n \). The power series for \( b_n \) will converge for \( |x|<2 \).

Step 3: Conclusion.
The correct answer is (A).