To determine the radius of convergence and interval of convergence for the series

$\sum_{n=1}^\infty \frac{(x-5)^n}{n^2},$

we'll use the ratio test and analyze the convergence:

Step 1: Ratio Test

The general term of the series is: $a_n = \frac{(x-5)^n}{n^2}.$

Using the ratio test: $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{(x-5)^{n+1}}{(n+1)^2}}{\frac{(x-5)^n}{n^2}} \right|.$

Simplify: $\left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(x-5)^{n+1}}{(n+1)^2} \cdot \frac{n^2}{(x-5)^n} \right| = \lim_{n \to \infty} \left| \frac{(x-5) \cdot n^2}{(n+1)^2} \right|.$

Factor and simplify further: $\lim_{n \to \infty} \left| \frac{n^2}{(n+1)^2} \cdot (x-5) \right| = \lim_{n \to \infty} \left| \frac{1}{(1+\frac{1}{n})^2} \cdot (x-5) \right|.$

As $n \to \infty$, $(1 + \frac{1}{n})^2 \to 1$, so the limit becomes: $\lim_{n \to \infty} \left| x-5 \right| < 1.$

Step 2: Radius of Convergence

From the inequality: $|x-5| < 1.$

The radius of convergence is: $R = 1.$

Step 3: Interval of Convergence

The inequality $|x-5| < 1$ gives: $4 < x < 6.$

Now check the endpoints $x = 4$ and $x = 6$ separately:

At $x = 4$:

The series becomes: $\sum_{n=1}^\infty \frac{(4-5)^n}{n^2} = \sum_{n=1}^\infty \frac{(-1)^n}{n^2}.$ This is the alternating series $\sum (-1)^n / n^2$, which converges by the alternating series test and absolute convergence (p-series with $p=2 > 1$).

At $x = 6$:

The series becomes: $\sum_{n=1}^\infty \frac{(6-5)^n}{n^2} = \sum_{n=1}^\infty \frac{1^n}{n^2} = \sum_{n=1}^\infty \frac{1}{n^2}.$ This is a p-series with $p=2 > 1$, so it converges.

Final Answer:

The interval of convergence is: $[4, 6].$

The radius of convergence is: $R = 1.$

Would you like further clarification or additional examples? Here are 5 related questions for practice:

1. What is the radius of convergence for $\sum_{n=1}^\infty \frac{(x-3)^n}{2^n}$?
2. How does the ratio test differ from the root test in finding the radius of convergence?
3. For the series $\sum_{n=1}^\infty \frac{(-1)^n (x+2)^n}{n^3}$, find the interval of convergence.
4. What happens if the series contains factorial terms, such as $\sum_{n=1}^\infty \frac{(x-1)^n}{n!}$?
5. How do you test for absolute convergence versus conditional convergence at the endpoints?

Tip: Always test the endpoints of an interval of convergence separately, as their behavior often differs from the open interval.