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$\displaystyle \lim_{n \to \infty}\bigg( \dfrac{1}{2\cdot 3}+ \frac{1}{3\cdot 4}+...+ \frac{1}{(n+1)(n+2)} \bigg)=\\ \\ \\ = \lim_{n \to \infty} \bigg( \frac{3-2}{2\cdot 3} + \frac{4-3}{3\cdot 4} +...+ \frac{n+2-(n+1)}{(n+1)(n+2)} \bigg)=\\ \\ \\ = \lim_{n \to \infty} \bigg(\frac{1}{2}-\frac{1}{3} +\frac{1}{3}-\frac{1}{4} +...+ \frac{1}{n+1} -\frac{1}{n+2} \bigg)=\\ \\ \\ = \lim_{n \to \infty} \bigg(\frac{1}2}- \frac{1}{n+2}\bigg)=\frac{1}{2}-0=\frac{1}2}$