Answer:

1. (a) \(a=2\) and \(b=1\), so \(\sum_{r=2}^{n}\frac{1}{r^{2}-1}=\frac{3}{4}-\frac{2n+1}{2n(n+1)}\).
2. (b) \(\sum_{r=2}^{\infty}\frac{1}{r^{2}-1}=\frac{3}{4}\).
3. (c) The limit is \(\frac{1}{2}\).

1. Factorise the denominator: \(r^{2}-1=(r-1)(r+1)\). Write

   \(\frac{1}{r^{2}-1}=\frac{1}{(r-1)(r+1)}=\frac{A}{r-1}+\frac{B}{r+1}\).

   Multiplying by \((r-1)(r+1)\) gives \(1=A(r+1)+B(r-1)\). Substituting \(r=1\) gives \(A=\frac12\), and substituting \(r=-1\) gives \(B=-\frac12\). Hence

   \(\frac{1}{r^{2}-1}=\frac12\left(\frac{1}{r-1}-\frac{1}{r+1}\right)\).

   Therefore

   \(\sum_{r=2}^{n}\frac{1}{r^{2}-1}=\frac12\left(\left(1-\frac13\right)+\left(\frac12-\frac14\right)+\left(\frac13-\frac15\right)+\cdots+\left(\frac{1}{n-1}-\frac{1}{n+1}\right)\right)\).

   Most terms cancel, leaving

   \(\sum_{r=2}^{n}\frac{1}{r^{2}-1}=\frac12\left(1+\frac12-\frac1n-\frac{1}{n+1}\right)\).

   So

   \(\sum_{r=2}^{n}\frac{1}{r^{2}-1}=\frac34-\frac12\left(\frac1n+\frac{1}{n+1}\right)=\frac34-\frac{2n+1}{2n(n+1)}\).

   Comparing this with \(\frac34-\frac{an+b}{2n(n+1)}\), we get \(a=2\) and \(b=1\).

2. From part (a),

   \(\sum_{r=2}^{n}\frac{1}{r^{2}-1}=\frac34-\frac{2n+1}{2n(n+1)}\).

   As \(n\to\infty\), \(\frac{2n+1}{2n(n+1)}\to 0\). Therefore

   \(\sum_{r=2}^{\infty}\frac{1}{r^{2}-1}=\frac34\).

3. Write the required finite sum as a difference of two sums:

   \(\sum_{r=n+1}^{2n}\frac{n}{r^{2}-1}=n\sum_{r=n+1}^{2n}\frac{1}{r^{2}-1}=n\left(\sum_{r=2}^{2n}\frac{1}{r^{2}-1}-\sum_{r=2}^{n}\frac{1}{r^{2}-1}\right)\).

   Using part (a) with upper limit \(2n\),

   \(\sum_{r=2}^{2n}\frac{1}{r^{2}-1}=\frac34-\frac{4n+1}{4n(2n+1)}\).

   Also,

   \(\sum_{r=2}^{n}\frac{1}{r^{2}-1}=\frac34-\frac{2n+1}{2n(n+1)}\).

   Hence

   \(\sum_{r=n+1}^{2n}\frac{n}{r^{2}-1}=n\left(\frac{2n+1}{2n(n+1)}-\frac{4n+1}{4n(2n+1)}\right)\).

   So

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

   Taking the limit as \(n\to\infty\),

   \(\frac{2n+1}{2(n+1)}\to 1\), while \(\frac{4n+1}{4(2n+1)}\to \frac12\). Therefore the required limit is

   \(1-\frac12=\frac12\).