To determine the radius of convergence $R$ for the series

$\sum_{n=1}^{\infty} \frac{n^4 x^n}{17^n},$

we'll use the Root Test or Ratio Test.

Step-by-Step Solution

1. Rewrite the Series Terms: The general term of the series is: $a_n = \frac{n^4 x^n}{17^n}.$

2. Apply the Ratio Test: According to the Ratio Test, the radius of convergence $R$ can be found by examining the limit: $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|.$ If this limit exists and equals $\frac{1}{R}$, then we can solve for $R$.

3. Calculate $\frac{a_{n+1}}{a_n}$: Substitute $a_{n+1} = \frac{(n+1)^4 x^{n+1}}{17^{n+1}}$ and $a_n = \frac{n^4 x^n}{17^n}$: $\frac{a_{n+1}}{a_n} = \frac{(n+1)^4 x^{n+1} / 17^{n+1}}{n^4 x^n / 17^n} = \frac{(n+1)^4}{n^4} \cdot \frac{x}{17}.$

4. Take the Limit as $n \to \infty$: $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left( \frac{(n+1)^4}{n^4} \cdot \frac{|x|}{17} \right).$ As $n \to \infty$, $\frac{(n+1)^4}{n^4} \to 1$, so: $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \frac{|x|}{17}.$

5. Set the Limit Less Than 1 for Convergence: For convergence, we need: $\frac{|x|}{17} < 1 \Rightarrow |x| < 17.$

6. Conclusion: The radius of convergence $R$ is therefore: $R = 17.$

Answer

The correct choice is:

• 17

Would you like more details, or do you have any questions about this solution?

Here are some additional questions to explore:

1. What is the significance of the Ratio Test in determining the convergence of a series?
2. How would the radius of convergence change if the series had a different base, say $10^n$ instead of $17^n$?
3. Can the Root Test also be applied to find the radius of convergence in this example?
4. What happens to the convergence of the series at the endpoints $x = \pm 17$?
5. How does the term $n^4$ in the numerator affect the convergence properties of the series?

Tip: The Radius of Convergence can always be interpreted as the range of values for which the series converges around the center point $x = 0$.