We are asked to determine whether the series $\sum_{n=1}^\infty (-1)^{n+1} \frac{\ln n}{n}$ converges.

1. Problem Description

We are asked to determine whether the series $\sum_{n=1}^\infty (-1)^{n+1} \frac{\ln n}{n}$ converges.

2. Solution Steps

We can apply the Alternating Series Test to determine if the series converges. The Alternating Series Test states that for a series of the form $\sum_{n=1}^\infty (-1)^{n+1} a_n$, where $a_n > 0$ for all $n$, the series converges if the following two conditions are met:

(1) $a_n$ is a decreasing sequence.

(2) $\lim_{n \to \infty} a_n = 0$.

In our case, $a_n = \frac{\ln n}{n}$.

First, let's check the limit condition:

$\lim_{n \to \infty} \frac{\ln n}{n}$.

This is of the form $\frac{\infty}{\infty}$, so we can use L'Hopital's Rule:

$\lim_{n \to \infty} \frac{\ln n}{n} = \lim_{n \to \infty} \frac{1/n}{1} = \lim_{n \to \infty} \frac{1}{n} = 0$.

Thus, the second condition is satisfied.

Now let's check if $a_n$ is a decreasing sequence. Consider the function $f(x) = \frac{\ln x}{x}$ for $x \geq 1$. We can find its derivative:

$f'(x) = \frac{x \cdot \frac{1}{x} - \ln x \cdot 1}{x^2} = \frac{1 - \ln x}{x^2}$.

For $x > e$, we have $\ln x > 1$, so $1 - \ln x < 0$. Thus, $f'(x) < 0$ for $x > e$, which means $f(x)$ is decreasing for $x > e$. Since $e \approx 2.718$, the sequence $\frac{\ln n}{n}$ is decreasing for $n \geq 3$. For $n=1$, $a_1 = \frac{\ln 1}{1} = 0$. For $n=2$, $a_2 = \frac{\ln 2}{2} \approx \frac{0.693}{2} \approx 0.346$. For $n=3$, $a_3 = \frac{\ln 3}{3} \approx \frac{1.099}{3} \approx 0.366$.

We have $a_1 < a_2 < a_3$. However, for $n \geq 3$, the sequence is decreasing. Since the sequence is eventually decreasing and the limit of the terms is zero, the alternating series converges.

3. Final Answer

The series converges.