Bonsoir,
1)[tex] \dfrac{1}{n} - \dfrac{1}{n+1} = \dfrac{n+1-n}{n(n+1)} =\dfrac{1}{n(n+1)}[/tex]
2)
[tex] \dfrac{1}{1*2} = \dfrac{1}{1} -\dfrac{1}{2}\\
\dfrac{1}{2*3} = \dfrac{1}{2} -\dfrac{1}{3}\\
\dfrac{1}{3*4} = \dfrac{1}{3} -\dfrac{1}{4}\\
...\\
\dfrac{1}{6*7} = \dfrac{1}{6} -\dfrac{1}{7}\\
\dfrac{1}{7*8} = \dfrac{1}{7} -\dfrac{1}{8}\\
\dfrac{1}{8*9} = \dfrac{1}{8} -\dfrac{1}{9}\\
\dfrac{1}{9*10} = \dfrac{1}{9} -\dfrac{1}{10}\\
--------on\ additionne\ membre\ \`a \ membre\\
\sum _{i=1} ^{9}\dfrac{1}{i*(i+1)} \\
= \dfrac{1}{1} -\dfrac{1}{10}\\
= \dfrac{9}{10} \\
[/tex]