Verify the identity $$\frac{2r-1}{r(r-1)}-\frac{2r+1}{r(r+1)}=\frac{2}{(r-1)(r+1)}$$
Hence, using the method of differences, prove that $$\sum_{r=2}^{n}\frac{2}{(r-1)(r+1)}=\frac{3}{2}-\frac{2n+1}{n(n+1)}$$
Deduce the sum of the infinite series $$\frac{1}{1\cdot 3}+\frac{1}{2\cdot 4}+\frac{1}{3\cdot 5}+...+\frac{1}{(n-1)(n+1)}+...$$
I have done the first two parts and have recognized that half times the sum of the second part will equal the sum of the series in part three. How do I proceed?
$$\begin{align} S&=\frac 1{1\cdot 3}+\frac 1{2\cdot 4}+\cdots+\frac 1{(r-1)(r+1)}\cdots\\ &=\lim_{n\to\infty}\frac 12\sum_{r=2}^n \frac 2{(r-1)(r+1)}\\ &=\frac 12\lim_{n\to\infty}\sum_{r=2}^n \frac 2{(r-1)(r+1)}\\ &=\frac 12 \lim_{n\to\infty}\left[\frac 32-\frac{2n+1}{n(n+1)}\right]\\ &=\frac 12 \left[\frac 32-\lim_{n\to\infty}\frac{2n+1}{n(n+1)}\right]\\ &=\frac 34 -\frac 12\left[\lim_{n\to\infty}\frac{2n+1}{n(n+1)}\right]\\ &=\frac 34 -\frac 12\left[\lim_{n\to\infty}\frac{2n}{n^2}\right]\\ &=\frac 34 -\frac 12\underbrace{\left[\lim_{n\to\infty}\frac 1n\right]}_0\\ &=\frac 34\qquad \blacksquare \end{align}$$