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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.

Prove that the series x/2 - x^2/3 + x^3/4 - ... + (-1)^(n+1) x^n/n is absolutely convergent for |x| < 1.

Explore the step-by-step proof that the series x/2 - x^2/3 + x^3/4 - ... + (-1)^(n+1) x^n/n is absolutely convergent for |x| < 1. Learn about key mathematical concepts such as the Ratio Test, power series, and logarithmic series. Suitable for advanced high school or early undergraduate students.

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