To solve this problem and find the interval of convergence, let’s break it down step by step:

Step 1: Express the general term $a_n$

The power series is: $\sum_{n=1}^\infty \frac{11^n}{(n+3)!}(x-8)^n$ The general term $a_n$ is: $a_n = \frac{11^n}{(n+3)!}(x-8)^n$

Step 2: Apply the Ratio Test

The ratio test states: $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$ Substituting $a_n$ and $a_{n+1}$: $a_{n+1} = \frac{11^{n+1}}{(n+4)!}(x-8)^{n+1}$ $\frac{a_{n+1}}{a_n} = \frac{\frac{11^{n+1}}{(n+4)!}(x-8)^{n+1}}{\frac{11^n}{(n+3)!}(x-8)^n}$ Simplify: $\frac{a_{n+1}}{a_n} = \frac{11^{n+1} (x-8)^{n+1} (n+3)!}{11^n (x-8)^n (n+4)!}$ Factor and simplify: $\frac{a_{n+1}}{a_n} = \frac{11 \cdot 11^n \cdot (x-8) \cdot (x-8)^n (n+3)!}{11^n \cdot (x-8)^n \cdot (n+4)(n+3)!}$ $\frac{a_{n+1}}{a_n} = \frac{11 (x-8)}{n+4}$

Take the limit as $n \to \infty$: $L = \lim_{n \to \infty} \left| \frac{11 (x-8)}{n+4} \right| = 0$

Since the ratio test results in $L = 0 < 1$ for all $x$, the series converges for all $x$.

Step 3: Interval of Convergence

Since the series converges for all $x$, the interval of convergence is: $(-\infty, \infty)$

Would you like more details on any step? Let me know!

Related Questions:

1. What is the radius of convergence for this series?
2. How does the factorial in the denominator affect the series’ convergence?
3. Can you apply the ratio test to a different type of series, such as a geometric series?
4. How do you confirm convergence at endpoints of the interval when necessary?
5. What happens if the ratio test gives $L = 1$?

Tip:

When simplifying factorial terms in ratio tests, cancel as many terms as possible to make calculations easier.