The problem asks us to show that the alternating series $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{\ln n}{n}$ converges and then estimate the error made by using the partial sum $S_9$ as an approximation to the sum $S$ of the series.

1. Problem Description

The problem asks us to show that the alternating series $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{\ln n}{n}$ converges and then estimate the error made by using the partial sum $S_9$ as an approximation to the sum $S$ of the series.

2. Solution Steps

First, we need to show that the series converges. We can use the Alternating Series Test. Let $a_n = \frac{\ln n}{n}$. We need to show that $a_n$ is decreasing for $n$ sufficiently large and that $\lim_{n \to \infty} a_n = 0$.

To show that $a_n$ is decreasing, we can consider the function $f(x) = \frac{\ln x}{x}$ for $x \geq 1$. Then $f'(x) = \frac{\frac{1}{x} \cdot x - \ln x \cdot 1}{x^2} = \frac{1 - \ln x}{x^2}$. Since $x^2 > 0$ for all $x$, the sign of $f'(x)$ depends on $1 - \ln x$.

$f'(x) < 0$ when $1 - \ln x < 0$, which means $\ln x > 1$, or $x > e$. Thus, $f(x)$ is decreasing for $x > e \approx 2.718$. Therefore, $a_n$ is decreasing for $n \geq 3$.

Next, we need to show that $\lim_{n \to \infty} a_n = 0$. We can use L'Hopital's Rule to evaluate this limit:

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

Since $a_n$ is decreasing for $n \geq 3$ and $\lim_{n \to \infty} a_n = 0$, the alternating series converges by the Alternating Series Test.

Now, we need to estimate the error made by using the partial sum $S_9$ as an approximation to the sum $S$ of the series. The error is bounded by the absolute value of the next term in the series, $a_{10}$.

$|S - S_9| \leq a_{10} = \frac{\ln 10}{10} \approx \frac{2.3026}{10} \approx 0.23026$.

3. Final Answer

The series $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{\ln n}{n}$ converges, and the error made by using $S_9$ as an approximation is at most $\frac{\ln 10}{10} \approx 0.23026$.

Final Answer: The final answer is $\boxed{0.23026}$