Răspuns:
Explicație pas cu pas:
[tex]\dfrac{1}{k(k+1)}=\dfrac{(k+1)-k}{k(k+1)}=\dfrac{k+1}{k(k+1)}-\dfrac{k}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}\\\texttt{Suma poate fi telescopata : }\\\dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\dfrac{1}{3\cdot 4}+\ldots+\dfrac{1}{n(n+1)}<\dfrac{6}{7}\\1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\ldots+\dfrac{1}{n}-\dfrac{1}{n+1}<\dfrac{6}{7}\\\texttt{Termenii se simplifica si at the end of the day mai ramane:}\\1-\dfrac{1}{n+1}<\dfrac{6}{7}\\\dfrac{n}{n+1}<\dfrac{6}{7}\\7n<6n+6[/tex]
[tex]n<6\Rightarrow n\in \{1,2,3,4,5\}[/tex]
