\(S=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+.......+\frac{1}{100^2}<\frac{1}{2}\)
\(S=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+........+\frac{1}{100^2}\)<\(\frac{1}{0.2}+\frac{1}{2.4}+\frac{1}{4.6}+.......+\frac{1}{98.100}\)
\(S=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}<\frac{50}{100}=\frac{49}{100}<\frac{1}{2}\)
Vậy \(\frac{49}{100}<\frac{1}{2}\)
Ta có 1/22<1/2*3
1/42<1/3*4
. . .
1/1002<1/99*100
=> S<1/2*3+1/3*4+...+1/99*100
=> S<1/2-1/3+1/3-1/4+...+1/99-1/100
=>S<1/2-1/100
=>S<49/100
Mà 49/100<1/2
=>S<1/2
S = 1/2^2 + 1/4^2 + 1/6^2 + ... + 1/100^2
suy ra: 4*S = 1 + 1/2^2 + 1/3^2 + ... + 1/50^2
có: 1/2^2 = 1/2*2 < 1/1*2
1/3^2 = 1/3*3 < 1/2*3
1/50^2 = 1/50*50 <1/49*50
1+ 1/2^2 + 1/3^2 + ... + 1/50^2 < 1 + 1/1*2 + 1/2*3 + ... +1/49*50
4*S< 1 + 1 - 1/2 + 1/2 - 1/3 + ... + 1/49 - 1/50
4*S < 2 - 1/50 = 99/50
S < 99/50 : 4 = 99/50 * 1/4 = 99/200 < 100/200 = 1/2
vậy S < 1/2 (đpcm)