Fie n un numar natural nenul.
[tex] \frac{1}{n(n+1)} = \frac{(n+1)-n}{n(n+1)} = \frac{n+1}{n(n+1)}- \frac{n}{n(n+1)}= \frac{1}{n}- \frac{1}{n+1} [/tex]
Deci [tex] \frac{1}{2*3}= \frac{1}{2}- \frac{1}{3} \\ \frac{1}{3*4}= \frac{1}{3}- \frac{1}{4} \\ ................... \\ \frac{1}{45*46}= \frac{1}{45}- \frac{1}{46} [/tex]
Insumand aceste relatii, obtinem: S=
[tex] \frac{1}{2}- \frac{1}{3}+ \frac{1}{3}- \frac{1}{4}+...+ \frac{1}{45}- \frac{1}{46}= \\ = \frac{1}{2}- \frac{1}{46}= \\ = \frac{23}{46}- \frac{1}{46}= \\ = \frac{22}{46}= \\ =\frac{11}{23} [/tex]
