To solve this problem, we need to prove that the given infinite series is absolutely convergent for $|x| < 1$.

The series is:

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

Step 1: Understand Absolute Convergence

To prove absolute convergence, we need to show that the series formed by the absolute values of the terms is convergent. The series formed by the absolute values of the terms is:

$\sum_{n=1}^{\infty} \left|\frac{x^n}{n}\right|$

Since $\left|(-1)^{n+1}\right| = 1$, the series simplifies to:

$\sum_{n=1}^{\infty} \frac{|x|^n}{n}$

Step 2: Compare with a Known Convergent Series

This is a power series of the form:

$\sum_{n=1}^{\infty} \frac{a_n}{n}, \text{ where } a_n = |x|^n$

We can recognize that this series is similar to the Taylor series for $-\ln(1-x)$, which converges for $|x| < 1$.

Step 3: Use the Ratio Test

We apply the Ratio Test to the series $\sum_{n=1}^{\infty} \frac{|x|^n}{n}$ to determine its convergence. For the Ratio Test, we calculate:

$\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \left|\frac{|x|^{n+1}/(n+1)}{|x|^n/n}\right| = \lim_{n \to \infty} \frac{|x|\cdot n}{n+1} = |x|\cdot \lim_{n \to \infty} \frac{n}{n+1} = |x|\cdot 1 = |x|$

The Ratio Test states that if this limit is less than 1, then the series converges. Since $|x| < 1$, the series $\sum_{n=1}^{\infty} \frac{|x|^n}{n}$ converges, which implies the original series converges absolutely for $|x| < 1$.

Conclusion

Thus, the series is absolutely convergent for $|x| < 1$.

Would you like any additional details, or do you have any questions?

Here are some related questions to consider:

1. How does the Ratio Test work for different types of series?
2. What is the difference between conditional and absolute convergence?
3. Can you prove the convergence of the original series using the Alternating Series Test?
4. How can the Taylor series for $-\ln(1-x)$ be derived and applied?
5. What happens to the series if $|x| \geq 1$?

Tip: Always check the radius of convergence when dealing with power series to understand the interval where the series converges.