2025 VJC P1 Q6

It is given that \(\sum_{r=2}^{n} \frac{1}{r(r^2-1)} = \frac{1}{4} + \frac{1}{2(n+1)} - \frac{1}{2n}\).

1. Show that \(\sum_{r=2}^{n} \frac{1}{r(r^2-1)}\) is less than \(\frac{1}{4}\).[2]
2. Give a reason why the series \(\sum_{r=2}^{\infty} \frac{1}{r(r^2-1)}\) converges, and write down its value.[2]
3. Find the smallest value of \(n\) for which \(\sum_{r=2}^{n} \frac{1}{r(r^2-1)}\) differs from \(\sum_{r=2}^{\infty} \frac{1}{r(r^2-1)}\) by less than \(0.0007\).[2]
4. Find \(\sum_{r=m+1}^{N} \frac{1}{(r-m)(r-m+1)(r-m+2)}\), where \(m\) and \(N\) are integers with \(N > m > 0\). (There is no need to express your answer as a single algebraic fraction.)[2]