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?
Best Answer
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*}