The problem you provided asks to determine whether the series

$\sum_{n=1}^{\infty} \frac{n(x-4)^n}{3^n}$

is absolutely convergent or divergent using the ratio test.

Solution

To apply the ratio test, we consider the absolute value of the general term of the series, $a_n = \frac{n |x-4|^n}{3^n}$.

1. Ratio Test Setup: The ratio test requires calculating the limit: $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$ where $a_n = \frac{n |x-4|^n}{3^n}$.

2. Compute $a_{n+1}$: $a_{n+1} = \frac{(n+1) |x-4|^{n+1}}{3^{n+1}} = \frac{(n+1) |x-4|^{n+1}}{3 \cdot 3^n}$

3. Set Up the Ratio: $\frac{a_{n+1}}{a_n} = \frac{(n+1) |x-4|^{n+1}}{3 \cdot 3^n} \cdot \frac{3^n}{n |x-4|^n} = \frac{(n+1) |x-4|}{3n}$

4. Take the Limit as $n \to \infty$: $L = \lim_{n \to \infty} \frac{(n+1) |x-4|}{3n} = \frac{|x-4|}{3} \lim_{n \to \infty} \frac{n+1}{n} = \frac{|x-4|}{3}$

5. Apply the Ratio Test Conclusion: The ratio test states that:

   • If $L < 1$, the series converges absolutely.
   • If $L > 1$, the series diverges.
   • If $L = 1$, the test is inconclusive.

   Therefore, the series will converge absolutely if $\frac{|x-4|}{3} < 1$ which simplifies to $|x-4| < 3$

6. Solution in Terms of $x$: This inequality implies: $-3 < x - 4 < 3$ or $1 < x < 7$

Conclusion

The series converges absolutely for $x$ in the interval $(1, 7)$.

Would you like further details or have any questions?

Here are 5 related questions to expand your understanding:

1. How would the ratio test differ if the series included a factorial in the numerator or denominator?
2. Can the ratio test determine convergence if $L = 1$?
3. How does absolute convergence differ from conditional convergence?
4. What would happen if the series was $\sum_{n=1}^{\infty} \frac{n^2 (x-4)^n}{3^n}$?
5. Why is it helpful to check for absolute convergence when dealing with power series?

Tip: The ratio test is particularly useful for series with terms involving powers, as it simplifies with ratios.