\(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{50^2}\)
Ta thấy \(\frac{1}{2^2}=\frac{1}{2.2}< \frac{1}{1.2};\frac{1}{3^2}=\frac{1}{3.3}< \frac{1}{2.3};......;\frac{1}{50^2}=\frac{1}{50.50}< \frac{1}{49.50}\)
Khi đó \(A=1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}=B\)
\(B=1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}=1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{49}-\frac{1}{50}\)
\(B=1+1-\frac{1}{50}=2-\frac{1}{50}< 2\)
Vì \(B< 2\)mà \(A< B\)nên \(A< 2\left(đpcm\right)\)
\(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}.\)Ta có:
\(\frac{1}{2^2}< \frac{1}{1.2}=1-\frac{1}{2}\); \(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\);...; \(\frac{1}{50^2}< \frac{1}{49.50}=\frac{1}{49}-\frac{1}{50}\)
=> \(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
=> \(A< 1+1-\frac{1}{50}\)
=> \(A< 2-\frac{1}{50}\)
=> \(A< 2\)
