Checking for convergence of $\sum_{n=0}^\infty\frac{(-1)^nn}{n^2+1}$ I have the following series $\sum_{n=0}^\infty\frac{(-1)^nn}{n^2+1}$ which I am checking for convergence. 
My working:
$\lim_{n\to\infty}|a_n|=0$ where $a_n=\frac{(-1)^nn}{n^2+1}$.
Doesn't that mean $\sum_{n=0}^\infty\frac{(-1)^nn}{n^2+1}$ is absolutely convergent? 
My book says it's conditionally convergent. What am I missing?

The sum of each
even-odd pair is
$\begin{array}\\
\dfrac{2n}{(2n)^2+1}-\dfrac{2n+1}{(2n+1)^2+1}
&=\dfrac{2n((2n+1)^2+1)-(2n+1)(4n^2+1)}{((2n)^2+1)((2n+1)^2+1)}\\
&=\dfrac{4n^2+2n-1}{(4n^2+1)(4n^2+4n+2)}\\
&\lt \dfrac1{8n^2}\\
\end{array}
$
and the sum of these converges.
Since
$\dfrac{2n}{(2n)^2+1}
\to 0$,
the sum of the whole series
converges.
