TA CÓ Vế trái <\(\frac{1}{1}+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}\)\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{49}-\frac{1}{50}\)
\(=2-\frac{1}{50}< 2\)
do đó VT <2(dpcm)
\(\frac{1}{1^2}=1\)
\(\frac{1}{2^2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3^2}< \frac{1}{2\cdot3}\)
\(.\) \(.\)
\(.\) \(.\)
\(.\) \(.\)
\(\frac{1}{50^2}< \frac{1}{49\cdot50}\)
Cộng vế với vế ta có \(:\)
\(C< 1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...........+\frac{1}{49\cdot50}\)
\(C< 1+\frac{49}{50}< \frac{50+49}{50}=\frac{99}{50}< \frac{100}{50}=2\)
Vậy \(C=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...........+\frac{1}{50^2}< 2\)
\(C=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...........+\frac{1}{50^2}< 1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+.......+\frac{1}{49\cdot50}\)
\(C=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+........+\frac{1}{50^2}< 1+1-\frac{1}{50}\)
\(C=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...........+\frac{1}{50^2}< 2-\frac{1}{50}< 2\)
\(\)Vậy \(C< 2\)( điều cần chứng minh )