How can I determine convergence for this equation: $\sum_{n=1}^\infty (-1)^{n+1}\frac{n}{n^2+4}$ Again, my equation is:
$$
\sum_{n=1}^\infty (-1)^{n+1}\frac{n}{n^2+4}
$$
I believe I'm supposed to use a basic comparison check, so am I able to ignore the -1 portion at the start of the equation? It just shifts the graph between a positive and negative value. Looking at the graph of this equation it seems to converge at zero. Ignoring the -1, I get this:
$$
a_n = {\frac {n}{n^2+4}}, b_n = \frac {n}{n^2}
$$
so then I would divide $a_n$ by $b_n$
$$
a_n = \frac {n}{n^2+4}* \frac {n^2}{n} = \frac {n^3}{n^3+4n}
$$
And that's as far as I get. I thought I could take the coefficients of the larger powers and if they are positive, then the series converges, but I'm still unsure if I can ignore the $-1$ as well as the $4n$ I end up with in the denominator. 
 A: Your series does not converge absolutely. That is, when we "ignore" the $(-1)^{n+1}$ factor, we have a divergent series. 
However, when we do not ignore the $(-1)^{n+1}$ factor, then you can use the alternating series test to show that your (alternating) series converges (conditionally).
A: To prove that
$\sum_{n=1}^{\infty} (-1)^n a_n$
converges 
it is sufficient
to show that
$a_n > a_{n+1}$
and
$\lim_{n \to \infty} a_n = 0$.
Go thee hence.
A: $\newcommand{\+}{^{\dagger}}%
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert #1 \right\vert}%
 \newcommand{\yy}{\Longleftrightarrow}$
$\ds{\sum_{n = 1}^{\infty}\pars{-1}^{n + 1}\,{n \over n^{2} + 4}:\ {\large ?}}$

\begin{align}
&\color{#0000ff}{\large\sum_{n = 1}^{\infty}\pars{-1}^{n + 1}\,{n \over n^{2} + 4}}
=
\sum_{n = 1}^{\infty}\pars{-1}^{n + 1}\,{1 \over 2}\,
\pars{{1 \over n - 2\ic} + {1 \over n + 2\ic}}
=
\Re\sum_{n = 1}^{\infty}{\pars{-1}^{n + 1} \over n - 2\ic}
\\[3mm]&=
\Re\sum_{n = 1}^{\infty}\pars{{-1 \over 2n - 2\ic} + {1 \over 2n - 1 - 2\ic}}
=
\Re\sum_{n = 1}^{\infty}{1 \over \pars{2n - 2\ic}\pars{2n - 1 - 2\ic}}
\\[3mm]&=
{1 \over 4}\Re\sum_{n = 1}^{\infty}{1 \over \pars{n - \ic}\pars{n - 1/2 - \ic}}
=
{1 \over 4}\Re\sum_{n = 0}^{\infty}
{1 \over \pars{n + 1 - \ic}\pars{n + 1/2 - \ic}}\qquad\qquad\qquad\pars{1}
\\[3mm]&=
{1 \over 4}\,\Re\bracks{%
{\Psi\pars{1 - \ic} - \Psi\pars{1/2 - \ic} \over \pars{1 - \ic} - \pars{1/2 - \ic}}}
=
\color{#0000ff}{\large{1 \over 2}
\Re\bracks{\Psi\pars{1 - \ic} - \Psi\pars{{1 \over 2} - \ic}}
\approx 0.0732}
\end{align}

$\Psi\pars{z}$ is the $\it\mbox{digamma function}$.
Numerically:

/* sum_0.cc 16-nov-2013
http://math.stackexchange.com/users/85343/felix-marin
http://math.stackexchange.com/questions/569758/how-can-i-determine-convergence-for-this-equation-sum-n-1-infty-1n1/569761#569761
*/
#include <cfloat>
#include <cmath>
#include <iomanip>
#include <iostream>
using namespace std;
typedef long double ldouble;
typedef unsigned long long ullong;
const ullong N=1000000ULL; // One million
const ullong TWO=2ULL; 
ldouble a_n(ullong n);

int main()
{
 ldouble sum=0;
 for ( ullong n = 1ULL ; n<=N ; ++n ) sum+=a_n(n);
 cout<<setprecision(4)<<"Sum = "<<sum<<endl;
 return 0;
}

ldouble a_n(ullong n)
{
 static const ldouble tol=1.05*pow(LDBL_EPSILON,1.0L/7.0L);
 static ldouble res,temp;

        if ( n>TWO ) {
           res=1.0L/n;
           temp=2.0L/n;
           if ( temp>tol ) res/=(1.0L + temp*temp);
           else {
              temp*=temp;
              res*=(1.0L - temp*(1.0L - temp));
           }
 } else if ( n==2ULL ) res=0.25L;
   else                res=( n==1ULL ) ? 0.2L:0;
 return ( (n%TWO)==0 ) ? (-res):res;
}

It yields:
$\tt Sum = 0.07321$
