Questions: Consider the power series [ sumn=1^infty fracn^3(x+4)^n6^n cdot n^11 / 3 ] Find the radius of convergence R. If it is infinite, type "infinity" or "inf". Answer: R= 6 What is the interval of convergence? Answer (in interval notation): (-10,2)

Solution Steps

To find the radius of convergence of a power series, we can use the ratio test. For a series of the form \(\sum a_n (x - c)^n\), the radius of convergence \(R\) is given by \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\). In this case, we identify \(a_n = \frac{n^3}{6^n \cdot n^{11/3}}\) and apply the ratio test to find \(R\).

1. Identify the general term \(a_n = \frac{n^3}{6^n \cdot n^{11/3}}\).
2. Apply the ratio test: \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\).
3. Simplify the expression to find the radius of convergence \(R\).

The general term of the power series is given by

\[ a_n = \frac{n^3}{6^n \cdot n^{11/3}} = \frac{1}{6^n \cdot n^{8/3}}. \]

To find the radius of convergence \(R\), we apply the ratio test:

\[ \frac{a_{n+1}}{a_n} = \frac{(n+1)^3}{6^{n+1} \cdot (n+1)^{11/3}} \cdot \frac{6^n \cdot n^{11/3}}{n^3}. \]

This simplifies to

\[ \frac{a_{n+1}}{a_n} = \frac{(n+1)^3 \cdot n^{11/3}}{6 \cdot n^3 \cdot (n+1)^{11/3}}. \]

Taking the limit as \(n\) approaches infinity, we find:

\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{(n+1)^3}{6 \cdot n^3} \cdot \frac{n^{11/3}}{(n+1)^{11/3}}. \]

This limit evaluates to

\[ \frac{1}{6}. \]

The radius of convergence \(R\) is given by

\[ R = \frac{1}{\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|} = \frac{1}{\frac{1}{6}} = 6. \]

Final Answer