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

Question

It is given that $\sum_{r=2}^{n} \frac{1}{r(r^2-1)} = \frac{1}{4} + \frac{1}{2(n+1)} - \frac{1}{2n}$. Show that $\sum_{r=2}^{n} \frac{1}{r(r^2-1)}$ is less than $\frac{1}{4}$.[2] Give a reason why the series $\sum_{r=2}^{\infty} \frac{1}{r(r^2-1)}$ converges, and write down its value.[2] 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] 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]

Accepted Answer

(b)$\frac{1}{4}$ (c)smallest value of n = 27 (d)$\frac{1}{4}+\frac{1}{2(N-m+2 )}-\frac{1}{2-(N-m+1)}$

2025 VJC P1 Q6

2025 VJC P1 Q6