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.
 A: This definitely converges by the alternating series test.  The AST asks that the unsigned terms decrease and have a limit of 0.  In your case, the terms $\frac{n}{n^2+1}$ do exactly that, so it converges. 
Now, which flavor of convergence?
If you take absolute values, the resulting series $\sum_n \frac{n}{n^2+1}$  diverges.  You can probably get this quickest by limit comparison:  terms are on the order of $1/n$.  Also, the integral test here is pretty fast because you can see the logarithm.  
To apply limit comparison, let's compare $\sum_n \frac{n}{n^2+1}$  to $\sum_n \frac{1}{n}$.  Dividing a term in the first by a term in the second gives
$$
(\frac{n}{n^2+1})/(\frac{1}{n}) = \frac{n^2}{n^2+1}.
$$
Taking the limit gives $L=1$.  Since $L>0$, both series "do the same thing."  Since $\sum_n \frac{1}{n}$ diverges, so does $\sum_n \frac{n}{n^2+1}$. 
Hence it converges conditionally because it converges, but the series of absolute values does not. 
A: 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.
A: In at lot of these sort of problems, it comes down to picking good bounds for the terms. For instance, increasing a denominator decreases the size of a fraction (assuming the denominator and numerator are positive, of course). You can increase the denominator by replacing 1 by n. This results in a smaller fraction. So n/(n2+1) <   n/(n2+n) . But  n/(n2+n)= 1/(n+1), and by the integral test, that diverges. Since the absolute values are larger than a series that diverges the absolute values diverge.
We can also increase the size of the term by decreasing the denominator by 1, getting n/n2 = 1/n.
The next trick it to take pairs of terms. We know that if n is even, then bn = n/(n2+1)  < 1/n. And if n is odd, then bn = -n/(n2+1)  < -1/(n+1). So  if n is even, then bn +  bn+1 < 1/n - 1/((n+1)+1) 
1/n - 1/((n+1)+1)  = 1/n - 1/(n+2) = ((n+2)-n)/n(n+2) = 2/n(n+2)
We can now use the integral test to show this converges.  Since the pairs of terms are smaller than a series that converges, it converges (note that although not all the terms in the original series are positive, all of the pairs of terms sum to a positive amount, therefore this test is valid).
