Răspuns:
∑ [tex]\frac{1}{k(k+1)}[/tex] = ∑ [tex]\frac{k+1-k}{k(k+1)}[/tex] = ∑ [tex](\frac{k+1}{k(k+1)} - \frac{k}{k(k+1)})[/tex] = ∑ [tex](\frac{1}{k} - \frac{1}{k+1} )[/tex] [tex]= (\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} -\frac{1}{3} ) + (\frac{1}{3} - \frac{1}{4}) + ... +(\frac{1}{n} - \frac{1}{n+1} ) = \frac{1}{1} - \frac{1}{n+1} =\frac{n+1-1}{n+1} = \frac{n}{n+1}[/tex]
