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?
 A: The alternating series test shows that the original series converges, as you have shown. However, for $n\ge1$,
$$\frac n{n^2+1}=\frac1{n+1/n}\ge\frac1{n+1}$$
and so $\sum_{n=0}^\infty|a_n|>\sum_{n=1}^\infty\frac1{n+1}$, which diverges. The original series thus only conditionally, not absolutely, converges.
A: For a non-negative sequence $(a_{n})$, if $a_{n}\rightarrow 0$, we cannot conclude that $\displaystyle\sum_{n}a_{n}<\infty$. A typical example is $a_{n}=\dfrac{1}{n}$.
Note that $\dfrac{n}{n^{2}+1}$ is decreasing for $n\geq 2$, so alternating series test applies.
Meanwhile, it is not absolutely convergent because of the inequality that $\dfrac{n}{n^{2}+1}\geq\dfrac{n}{n^{2}+n^{2}}=\dfrac{1}{2}\dfrac{1}{n}$, so $\displaystyle\sum_{n}\dfrac{n}{n^{2}+1}\geq\dfrac{1}{2}\sum_{n}\dfrac{1}{n}=\infty$.
A: I guess you found $\lim_{n\to\infty} |a_n| = 0$ to use Alternating Series Test(AST). You should also verify that $|a_n|$'s are monotone decreasing. This test is used to show conditional convergence, does not imply absolute convergence. Here is the most common example to illustrate;
\begin{equation}
\sum_{n=1}^{\infty}\frac{(-1)^n}{n} \text{     is convergent due to Alternating Series Test.}
\end{equation}
On the other hand;
\begin{equation}
\sum_{n=1}^{\infty}\frac{1}{n} \text{   is divergent, called Harmonic Series.}
\end{equation}

To explaing further; conditional convergence means that $\sum a_n$ is convergent, but $\sum |a_n|$ is divergent. AST is a test for series of the form $\sum (-1)^na_n$, $a_n\geq 0$, and has no claim on $\sum a_n$.
A: 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.
