Summation of series of $2/(r-1)(r+1)$ using the method of differences 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?

$$\begin{align}
S&=\frac 1{1\cdot 3}+\frac 1{2\cdot 4}+\cdots+\frac 1{(r-1)(r+1)}\cdots\\
&=\lim_{n\to\infty}\frac 12\sum_{r=2}^n \frac 2{(r-1)(r+1)}\\
&=\frac 12\lim_{n\to\infty}\sum_{r=2}^n \frac 2{(r-1)(r+1)}\\
&=\frac 12 \lim_{n\to\infty}\left[\frac 32-\frac{2n+1}{n(n+1)}\right]\\
&=\frac 12 \left[\frac 32-\lim_{n\to\infty}\frac{2n+1}{n(n+1)}\right]\\
&=\frac 34 -\frac 12\left[\lim_{n\to\infty}\frac{2n+1}{n(n+1)}\right]\\
&=\frac 34 -\frac 12\left[\lim_{n\to\infty}\frac{2n}{n^2}\right]\\
&=\frac 34 -\frac 12\underbrace{\left[\lim_{n\to\infty}\frac 1n\right]}_0\\
&=\frac 34\qquad \blacksquare \end{align}$$
