Considere, $$f(n)=\frac{1}{(2n-1) \cdot (2n-1+2)}$$
donde $n$ es un número natural
$$\sum_{n=1}^\infty f(n)=\sum_{n=1}^\infty\frac{1}{(2n-1) \cdot (2n+1)}$$
Déjalo, $$\sum_{n=1}^{\infty} f(n) = S$$
es decir $$S=\sum_{n=1}^\infty\frac{1}{(2n-1)\cdot(2n+1)}$$ es decir $$S=\sum_{n=1}^\infty(\frac{1}{2})(\frac{1}{2n-1} - \frac{1}{2n+1})$$
es decir $$S=(\frac{1}{2})\left(\sum_{n=1}^\infty(\frac{1}{2n-1} - \frac{1}{2n+1})\right)$$
es decir $$S=\lim_{n \to \infty}\left((\frac{1}{2})(\frac{1}{1} - \frac{1}{3}) + (\frac{1}{2})(\frac{1}{3} - \frac{1}{5}) + (\frac{1}{2})(\frac{1}{5} - \frac{1}{7}) + \cdots + (\frac{1}{2})(\frac{1}{2n-1} - \frac{1}{2n+1})\right)$$
es decir $$S=\lim_{n \to \infty}(\frac{1}{2})\left((\frac{1}{1} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{5}) + (\frac{1}{5} - \frac{1}{7}) + \cdots + (\frac{1}{2n-1} - \frac{1}{2n+1})\right)$$
es decir $$S=\lim_{n \to \infty}(\frac{1}{2})\left(\frac{1}{1} - \frac{1}{2n+1}\right)$$
es decir $$S=(\frac{1}{2})\left(\frac{1}{1} - \lim_{n \to \infty}(\frac{1}{2n+1})\right)$$
es decir $$S=(\frac{1}{2})\left(\frac{1}{1} - 0\right)$$
es decir $$S=(\frac{1}{2})\left(\frac{1}{1}\right)$$
es decir $$S=\frac{1}{2}$$