Question: Consider the following series
$$\sum\_{n=1}^{\infty}\dfrac{\log n}{n(n-1)}$$
I have tried to use the ratio test, but then I would get
$$\dfrac{(n-1)\log(n+1)}{(n+1)\log n}$$
And taking the limit as $n \to \infty$ would yield 1 so I don't think it would help.

Answer: 
$\dfrac {4x}{x+1}=\dfrac{4x+4-4}{x+1}=4-\dfrac {4}{x+1}$

When $x$ tends to $0$ then $4-\dfrac {4}{x+1}$ tends to $0$. When $x$ tends to $+\infty$ then $4-\dfrac {4}{x+1}$ tends to $4$.

So, because your function is continuous your image is $(0,4)$.

Alternative approach:

$\dfrac {y}{4-y} >0$

This is possible if and only if $y>0$ and $4-y>0$ or if $y<0$ and $4-y<0$. See what case out of these two cases is possible.
$\frac{4x}{x+1}
=\frac{4x+4-4}{x+1}
=\frac{4(x+1)-4}{x+1}
=4-\frac{4}{x+1}
$

Now, the ranges.

$\begin{array}\\
x+1: 
&(0, \infty)\to (1, \infty)\\
\frac1{x}:
&(1, \infty) \to (0, 1)\\
4x:
&(0, 1)\to (0, 4)\\
4-x:
&(0, 4) \to 0, 4)\\
\end{array}
$
So the range is
$(0, 4)$.