Ta có :
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}< 1\)
\(\Rightarrow A< 1\)
ta có:
A=\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+....+\(\frac{1}{100^2}\)< B=\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+.....+\(\frac{1}{99.100}\)
B=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+........+\(\frac{1}{99}\)-1/100
B=1-1/100=99/100<1
Vì a<b mà B lại bé hơn 1 =>A<1
Do 1/2^2 + 1/3^2 + 1/4^2 +...+ 1/100^2 < 1/1.2 + 1/2.3 + 1/3.4 +...+ 1/99.100 (1)
Mà 1/2 + 1/3 + 1/4 +...+ 1/100= 1-1/2 + 1/2 - 1/3 + 1/3 - 1/4 +...+ 1/99 - 1/100
= 1-1/100 = 99/100 <1 (2)
Từ 1 và 2 => 1/2^2 + 1/3^2 + 1/4^2 +...+ 1/100^2 < 1
hay A < 1
Ta có : \(\frac{1}{2^2}\)< \(\frac{1}{1.2}\)
\(\frac{1}{3^2}\)< \(\frac{1}{2.3}\)
................
\(\frac{1}{100^2}\)<\(\frac{1}{99.100}\)
=> \(\frac{1}{2^2}\)+ \(\frac{1}{3^2}\)+ ...... + \(\frac{1}{100^2}\)< \(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+.......+ \(\frac{1}{99.100}\)
=> A < \(\frac{1}{1}\)- \(\frac{1}{2}\)+ \(\frac{1}{2}\)- \(\frac{1}{3}\)+ ..... + \(\frac{1}{99}\)- \(\frac{1}{100}\)
=> A < 1 - \(\frac{1}{100}\)<1
Vì 1 - \(\frac{1}{100}\)< 1
=> A < 1
Vậy A < 1
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(\frac{1}{2^2}+\frac{1}{3^2} +...+\frac{1}{100^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< 1-\frac{1}{100}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< 1\)