Oblicz wartość sumy

[M4V3R1CK]

Polecenie j/w.

\(\displaystyle{ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + ... + \frac{1}{1990 \cdot 1991}}\)

[tometomek91]

\(\displaystyle{ \sum_{n=1}^{1991} \frac{1}{n(n+1)}=\frac{n}{n+1}}\)

[iLoNkAaa]

tak to wygląda w teorii:
\(\displaystyle{ \frac{1}{n(n+1)}= \frac{1}{n}- \frac{1}{n+1}}\)
a tak w praktyce:
\(\displaystyle{ \frac{1}{1}- \frac{1}{2}+ \frac{1}{2}- \frac{1}{3}+ \frac{1}{3}- \frac{1}{4}+... + \frac{1}{1990}- \frac{1}{1991}=1- \frac{1}{1991}= \frac{1990}{1991}}\)