We are asked to determine whether the series $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{n}{n^2+1}$ converges or diverges. Furthermore, if it converges, we want to determine if it converges absolutely or conditionally.

1. Problem Description

We are asked to determine whether the series $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{n}{n^2+1}$ converges or diverges. Furthermore, if it converges, we want to determine if it converges absolutely or conditionally.

2. Solution Steps

Let $a_n = \frac{n}{n^2+1}$.

First, let us examine if the series $\sum_{n=1}^{\infty} |(-1)^{n+1} \frac{n}{n^2+1}| = \sum_{n=1}^{\infty} \frac{n}{n^2+1}$ converges absolutely.

We use the Limit Comparison Test with $b_n = \frac{1}{n}$.

$\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{n}{n^2+1}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{n^2}{n^2+1} = \lim_{n \to \infty} \frac{1}{1+\frac{1}{n^2}} = 1$.

Since $\sum_{n=1}^{\infty} \frac{1}{n}$ is a divergent harmonic series, by the Limit Comparison Test, $\sum_{n=1}^{\infty} \frac{n}{n^2+1}$ also diverges.

Therefore, the series $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{n}{n^2+1}$ does not converge absolutely.

Now, let us check if the alternating series $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{n}{n^2+1}$ converges conditionally using the Alternating Series Test.

For the alternating series test to apply, we must show that $a_n = \frac{n}{n^2+1}$ is decreasing and $\lim_{n \to \infty} a_n = 0$.

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

To check if $a_n$ is decreasing, we can consider the derivative of the continuous function $f(x) = \frac{x}{x^2+1}$.

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

For $x>1$, $f'(x) < 0$.

Thus, $a_n = \frac{n}{n^2+1}$ is decreasing for $n>1$.

Therefore, by the Alternating Series Test, the series $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{n}{n^2+1}$ converges.

Since the series converges but does not converge absolutely, it converges conditionally.

3. Final Answer

The series converges conditionally.