<p>It is given that&nbsp;\(\sum_{r=2}^{n} \frac{1}{r(r^2-1)} = \frac{1}{4} + \frac{1}{2(n+1)} - \frac{1}{2n}\).</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Show that&nbsp;\(\sum_{r=2}^{n} \frac{1}{r(r^2-1)}\) is less than \(\frac{1}{4}\).</span><span class="inline-marks">[2]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Give a reason why the series \(\sum_{r=2}^{\infty} \frac{1}{r(r^2-1)}\) converges, and write down its value.</span><span class="inline-marks">[2]</span></div></li><li><div class="list-item-content"><span class="list-item-text">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 \(0.0007\).</span><span class="inline-marks">[2]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find \(\sum_{r=m+1}^{N} \frac{1}{(r-m)(r-m+1)(r-m+2)}\), where&nbsp;\(m\) and \(N\) are integers with \(N &gt; m &gt; 0\). (There is no need to express your answer as a single algebraic fraction.)</span><span class="inline-marks">[2]</span></div></li></ol>

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 \(\frac{1}{4}\).

Give a reason why the series \(\sum_{r=2}^{\infty} \frac{1}{r(r^2-1)}\) converges, and write down its value.

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 \(0.0007\).

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.)

2025 VJC P1 Q6

2025 VJC P1 Q6