Ta có : \(\frac{1}{2^2}<\frac{1}{1\cdot2}\)
\(\frac{1}{3^2}<\frac{1}{2\cdot3}\)
...
\(\frac{1}{100^2}<\frac{1}{99\cdot100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+.......+\frac{1}{100^2}<\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+.......+\frac{1}{99\cdot100}\)
Ta có : \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+.......+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.......+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}<1\)
\(\Rightarrow\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+.......+\frac{1}{99\cdot100}<1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+.......+\frac{1}{100^2}<1\Rightarrow1+\frac{1}{2^2}+\frac{1}{3^2}+.......+\frac{1}{100^2}<2\)
Ta có
1 + 1/2^2 + 1/3^2+.....+1/100^2 = 1,634939
=)) 1,634939 < 2
cái này thì để tui trả lời cho yên tâm
Ta cho tổng trên là A
A=\(1+\frac{1}{2^2}+\frac{1}{3^2}+..........+\frac{1}{100^2}\)
Ax2=A2
A2=\(2+\frac{1}{2^2}+\frac{1}{3^2}+........+\frac{1}{100^2}\)
A2-2=A
suy ra A=\(2-\frac{1}{2^2}+\frac{1}{3^2}+......\frac{1}{100^2}\)
Mà \(2-\frac{1}{2^2}+\frac{1}{3^2}+......\frac{1}{100^2}\)<2
Nên tổng trên bé hơn 2