[tex]\text{Tinem cont de faptul ca:}\\\dfrac{1}{k(k+1)}=\dfrac{k+1-k}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}\\\\\dfrac{1}{2\cdot 3}+\dfrac{1}{3\cdot 4}+\dfrac{1}{4\cdot 5}+\ldots +\dfrac{1}{2015\cdot 2016}<\dfrac{1}{2}\\\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\ldots+\dfrac{1}{2015}-\dfrac{1}{2016}<\dfrac{1}{2}\\\text{Termenii se reduc:}\\\dfrac{1}{2}-\dfrac{1}{2016}<\dfrac{1}{2}\\-\dfrac{1}{2016}<0(Adevarat)[/tex]
