Bài 1:
Ta có:
$\frac{1}{n^2+3n+2}=$ $\frac{1}{(n+1)(n+2)}=$ $\frac{1}{n+1}-$ $\frac{1}{n+2}$
$\frac{1}{6}=$ $\frac{1}{2.3}=$ $\frac{1}{2}-$ $\frac{1}{3}$
$\frac{1}{12}=$ $\frac{1}{3.4}=$ $\frac{1}{3}-$ $\frac{1}{4}$
$\frac{1}{20}=$ $\frac{1}{4.5}=$ $\frac{1}{4}-$ $\frac{1}{5}$
$\frac{1}{n^2+3n+2}=$ $\frac{1}{(n+1)(n+2)}=$ $\frac{1}{n+1}+$ $\frac{1}{n+2}$
$\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{n^2+3n+2}=\frac{1}{2}-\frac{1}{n+2}<\frac{1}{2}$
Bài 2:
Ta có: $\frac{1}{a(a+1)}=\frac{1}{a}-\frac{1}{a+1}$
⇒ $S=(\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+...+(\frac{1}{99}-$$\frac{1}{100})=1-\frac{1}{100}=$ $\frac{99}{100}$
