Answer
Function satisfies conditions for integral test
Series diverges
Work Step by Step
$\dfrac {1}{4}+\dfrac {2}{7}+\dfrac {3}{12}+\ldots \dfrac {n}{n^{2}+3}=\sum ^{\infty }_{1}\dfrac {n}{n^{2}+3}$
In order integral test can be applied $f\left( x\right) =\dfrac {x}{x^{2}+3} $ must be positive continuous and decreasing for $x>1$
We see that function is positive and continuous
$f'\left( x\right) =\dfrac {x^{2}+3-2x^{2}}{\left( x^{2}+3\right) ^{2}}=\dfrac {3-x^{2}}{\left( x^{2}+1\right) } $
the function is decreasing
So integral test can be applied
$$ \int ^{\infty }_{1}\dfrac {x}{x^{2}+3}=\dfrac {1}{2}\ln \left( x^{2}+3\right) ]^{\infty }_{1}=\infty $$
So the series diverges
