We are asked to analyze the series $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{n}{n^2+1}$. First, we need to show that this alternating series converges. Then, we need to estimate the error made by using the partial sum $S_9$ as an approximation to the sum $S$ of the series.

1. Problem Description

We are asked to analyze the series $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{n}{n^2+1}$. First, we need to show that this alternating series converges. Then, we need to estimate the error made by using the partial sum $S_9$ as an approximation to the sum $S$ of the series.

2. Solution Steps

To show that the alternating series $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{n}{n^2+1}$ converges, we need to verify two conditions:

(1) $b_n = \frac{n}{n^2+1} > 0$ for all $n \geq 1$.

(2) $b_n$ is decreasing.

(3) $\lim_{n \to \infty} b_n = 0$

Condition (1) is clearly satisfied since $n$ and $n^2+1$ are both positive for $n \geq 1$.

To verify condition (2), we consider the function $f(x) = \frac{x}{x^2+1}$ for $x \geq 1$. We want to show that $f'(x) < 0$.

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

Since $x \geq 1$, $1-x^2 \leq 0$. Also $(x^2+1)^2 > 0$. Thus $f'(x) \leq 0$ for $x \geq 1$. This means that the sequence $b_n = \frac{n}{n^2+1}$ is decreasing.

To verify condition (3), we need to show that $\lim_{n \to \infty} \frac{n}{n^2+1} = 0$.

$\lim_{n \to \infty} \frac{n}{n^2+1} = \lim_{n \to \infty} \frac{n/n^2}{(n^2+1)/n^2} = \lim_{n \to \infty} \frac{1/n}{1+1/n^2} = \frac{0}{1+0} = 0$.

Therefore, the alternating series converges by the Alternating Series Test.

Now, to estimate the error made by using the partial sum $S_9$ as an approximation to the sum $S$ of the series, we can use the Alternating Series Estimation Theorem. This theorem states that $|S - S_n| \leq b_{n+1}$.

In our case, $n=9$, so we have $|S - S_9| \leq b_{10} = \frac{10}{10^2+1} = \frac{10}{100+1} = \frac{10}{101}$.

Therefore, the error made by using $S_9$ to approximate $S$ is at most $\frac{10}{101}$.

3. Final Answer

The series converges by the Alternating Series Test. The error is at most $\frac{10}{101}$.

Final Answer: The final answer is $\boxed{\frac{10}{101}}$