Here, $u_n = \dfrac {x^n}{n(n+1)} \ \text{ and } u_{n+1} = \dfrac {x^{n+1}}{(n+1)(n+2)}$

$\therefore \dfrac {u_n}{u_{n+1}} = \dfrac {n+2}{n} \dfrac {1}{x} $

$\displaystyle \lim_{n\to \infty} \dfrac {u_n}{u_{n+1}} = \lim_{n\to \infty} \left (1+ \dfrac {2}{n} \right )\dfrac {1}{2} = \dfrac {1}{x}$

$ \therefore \sum \ u_n$ is convergent if (\dfrac {1}{x} > 1) i.e. (x<1\) and divergent if \(\dfrac {1}{x}<1\) i.e. \(x>1) the test fails if $x=1$

But when x=1, $u_n = \dfrac {1}{n(n+1)} = \dfrac {1}{n} - \dfrac {1}{n+1}$

Hence $s_n = \left ( \dfrac {1}{1}- \dfrac {1}{2} \right ) + \left ( \dfrac {1}{2} - \dfrac {1}{3} \right )+ \left ( \dfrac {1}{3} - \dfrac {1}{4} \right ) + \cdots \ + \left ( \dfrac {1}{n} - \dfrac {1}{n+1} \right )=1 - \dfrac {1}{n+1}$

$\displaystyle \lim_{n\to \infty} \ s_n=1$

Hence,

$ \therefore \sum \ u_n$ is convergent if $x\le 1$ and divergent if (x>1)