How to find the partial sum of a given series? On my last exam there was the question if the series $\sum_{n=2}^{\infty}\frac{1}{(n-1)n(n+1)}$ converges and which limit it has. During the exam and until now, I am not able to solve it. I tried partial fraction decomposition, telescoping sum, etc. But I am not able to find the partial sum formula (Wolfram|Alpha):
$$
\sum_{n=2}^{m}\frac{1}{(n-1)n(n+1)} = \frac{m^2+m-2}{4m(m+1)}.
$$
Could somebody push me in the right direction? Is there any trick or scheme how to find partial sum formulas for given series?
 A: So let's try partial fraction decomposition. Writing
$$
  \frac 1{(n-1)n(n+1)} = \frac a{n-1} + \frac bn + \frac c{n+1}
$$
we obtain
$$
   1 = a(n^2 + n) + b(n^2 - 1) + c(n^2 - n)
$$
and therefore
\begin{align*}
   1 &= -b\\
   0 &= a - c\\
   0 &= a + b + c.
\end{align*}
This gives $b = -1$, $a = c = \frac 12$. Hence
\begin{align*}
  \sum_{n=2}^m \frac 1{(n-1)n(n+1)} 
   &= \sum_{n =2}^m \frac 1{2(n-1)} - \sum_{n=2}^m \frac 1n + \sum_{n=2}^m \frac 1{2(n+1)}\\
   &= \frac 12 + \sum_{n=2}^{m-1} \frac 1{2n} - \sum_{n=2}^m \frac 1n 
         + \sum_{n=3}^m \frac 1{2n} + \frac 1{2(m+1)}\\
   &= \frac 12 + \frac 14 - \frac 12 - \frac 1m + \frac 1{2m} + \frac 1{2m+2}\\
   &= \frac 14 + \frac{-2(m+1) + m+1 + m}{2m(m+1)}\\
   &= \frac 14 + \frac{-1}{2m(m+1)}\\
   &= \frac{m(m+1) - 2}{4m(m+1)}.
\end{align*}
A: What did you try for a telescoping sum?  Your denominator here is the product of three successive terms (this is often called a rising or falling factorial, depending on which side you take as your baseline); this points to looking at a difference of terms that are of the same form but with denominators one degree less.  In particular, looking at $t_n=\dfrac{1}{n(n+1)}$ then $t_n-t_{n-1}$ $=\dfrac{1}{n(n+1)}-\dfrac{1}{(n-1)n}$ $=\dfrac1n\left(\dfrac1{n+1}-\dfrac1{n-1}\right)$ $=\dfrac1n\left(\dfrac{(n-1)-(n+1)}{(n-1)(n+1)}\right)$ $=\dfrac{-2}{(n-1)n(n+1)}$; in other words, $\dfrac{1}{(n-1)n(n+1)} = -\dfrac12(t_n-t_{n-1})$, and from here the telescopy should be fairly clear.
