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.
Best Answer
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.