Determine whether $\sum\limits_{n=1}^{\infty} (-1)^{n-1}(\frac{n}{n^2+1})$ is absolutely convergent, conditionally convergent, or divergent. 
Determine whether the series is absolutely convergent, conditionally convergent, or divergent.
$$\sum_{n=1}^{\infty} (-1)^{n-1}\left(\frac{n}{n^2+1}\right)$$

Here's my work:
$b_n = (\dfrac{n}{n^2+1})$
$b_{n+1} = (\dfrac{n+1}{(n+1)^2+1})$
$\lim\limits_{n \to \infty}(\dfrac{n}{n^2+1}) = \lim\limits_{n \to \infty}(\dfrac{1}{n+1/n})=0$
Then I simplified $b_n - b_{n+1}$ in hopes of showing that the sum would be greater than or equal to $0$, but I failed (and erased my work so that's why I haven't included it).
I know the limit of $|b_n|$ is also 0, and I can use that for testing conditional convergence there, but I would run into the same problem for the second half of the test.
I'm having trouble wrapping my head around tests involving absolute values, or more specifically when I have to simplify them.

One can apply the alternating series test directly, but an alternative approach is to take the difference with the series $\sum_{n=1}^\infty
(-1)^{n-1}/n$ which is well-known to converge conditionally.
Details: let $a_n=(-1)^{n-1} n/(n^2+1)$, $b_n=(-1)^{n-1}/n$ and $c_n=a_n-b_n$.
Then
$$c_n=(-1)^{n-1}\left(\frac n{n^2+1}-\frac1n\right)
=\frac{(-1)^n}{n(n^2+1)}.$$
Then $\sum_n b_n$ converges conditionally, and $\sum_n c_n$
converges absolutely, so $\sum_na_n=\sum_n(b_n+c_n)$ converges conditionally.
