Ta có:
1/1.2+1/2.3+1/3.4+...+1/49.50
=1-1/2+1/2-1/3+1/3-1/4+...+1/49-1/50
=1-1/50
Vì 1-1/50<1 nên 1/1.2+1/2.3+1/3.4+...+1/49.50<1
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)\(=1-\frac{1}{50}
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{1}-\frac{1}{50}\)
\(=\frac{50}{50}-\frac{1}{50}\)
\(=\frac{49}{50}\)
\(\Rightarrow\)Tổng trên < 1