[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},\forall k\in \mathbb{N}^*\\S_n=\dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\dfrac{1}{3\cdot 4}+\ldots+\dfrac{1}{n(n+1)}\\S_n=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\ldots +\dfrac{1}{n}-\dfrac{1}{n+1}\\\text{Termenii se reduc:}\\S_n=1-\dfrac{1}{n+1}\\\boxed{S_n=\dfrac{n}{n+1}}[/tex]