Question

The radius of convergence of the power series \[\sum_{n=1}^{\infty} \left(\frac{n^3}{4^n}\right) x^{5n}\] is

• A. 4
• B. \( \sqrt[5]{4} \)
• C. \( \frac{1}{4} \)
• D. \( \frac{1}{\sqrt[5]{4}} \)

Answer 1

The Correct Option is B

To determine the radius of convergence of the power series \(\sum_{n=1}^{\infty} \left(\frac{n^3}{4^n}\right) x^{5n}\), we can use the Root Test, which is suitable for series of the form \(\sum a_n x^n\). The Root Test provides that if the limit:

\(\lim_{n \to \infty} \sqrt[n]{|a_n|}\ = \frac{1}{R}\)

exists, then the radius of convergence \(R\) is computed from this formula.

1. In this problem, the series is \(\sum_{n=1}^{\infty} \left(\frac{n^3}{4^n}\right) x^{5n}\), thus, we can set:
   \(a_n = \frac{n^3}{4^n}\) and our series transforms into the form \(\sum_{n=1}^{\infty} a_n x^{5n}\).
2. To apply the Root Test, we calculate \(\lim_{n \to \infty} \sqrt[n]{|a_n|}\) as follows:
   \(\sqrt[n]{|a_n|} = \sqrt[n]{\frac{n^3}{4^n}} = \frac{(n^3)^{1/n}}{4}\)
3. We know that: \(\lim_{n \to \infty} (n^3)^{1/n} = \lim_{n \to \infty} n^{3/n} = 1\) (since \(\lim_{n \to \infty} n^{1/n} = 1\)).
4. Thus,
   \(\lim_{n \to \infty} \frac{(n^3)^{1/n}}{4} = \frac{1}{4}\)
5. Following the Root Test: \(\frac{1}{R} = \frac{1}{4}\) implies \(R = 4\)
6. We must consider the series as \(\sum a_n x^{5n}\) rather than a simple power series \(\sum a_n x^n\). Therefore, the transformation based on \(x^{5n}\) makes the actual radius: \(R_0 = R^{1/5} = 4^{1/5}\)
7. The correct calculation for the radius of convergence given the factor of 5 in \(x^{5n}\) leads us to: \(R_0 = \sqrt[5]{4}\)

Thus, the radius of convergence for the given power series is \(\sqrt[5]{4}\), making Option 2 the correct answer.