Ta có: A < \(\frac{1}{1^2}+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

Lại có: \(\frac{1}{1^2}+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}=1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

                                                                                 \(=1+\left(\frac{1}{1}-\frac{1}{50}\right)\)

                                                                                  \(=1+\frac{49}{50}\)

Mà 1+49/50<2 nên A<1+49/50<2

Vậy A<2

Ta có:
A=1+(1/2+1/3)+(1/4+1/5+1/6+1/7)+(1/8+1/9+......+1/15)+........+ (1/2^99+1/2^99+1+........+1/2^100-1)
(Có 99 nhóm) < 1+2.1/2+2^2.1/2^2+2^3.1/2^3+.....+2^99.1/2^99
=>1+1+1+.......+1 (100 số 1)=100
=>A1+1/2+2.1/2^2+2^2.1/2^3+2^3.1/2^4+.....+2^991/2^100-1-1/2^100 =1+1/2+1/2+1/2+1/2+........+1/2-1/2^100 (100 số 1/2)
=1+100.12-1/2^100
=50+1-1/2^100>50
=>A>50 (2)
Từ (1)và (2)=>50

Đặt A=1/2^2+1/3^2+1/4^2+...+1/50^2

       A<1/1*2+1/2*3+1/3*4+...+1/49*50

       A<1-1/2+1/2-1/3+1/3-1/4+...+1/49-1/50

      A<1-1/50<1

Vậy A<1

Ta có:\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{50^2}< \frac{1}{49.50}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}< 1\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}< 1\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< 1\left(đpcm\right)\)

\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+...+\(\frac{1}{50^2}\)<1

ta có \(\frac{1}{2^2}\)<\(\frac{1}{1.2}\)

       \(\frac{1}{3^2}\)<\(\frac{1}{2.3}\)

    ..........................

    \(\frac{1}{50^2}\)<\(\frac{1}{49.50}\)

ta được \(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+...+\(\frac{1}{49.50}\)

          =>1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-...-\(\frac{1}{49}\)+\(\frac{1}{49}\)-\(\frac{1}{50}\)

          =>1-\(\frac{1}{50}\)<1 nên\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+...+\(\frac{1}{50^2}\)<1

vậy ...........................

A=\(\frac{1}{1^2}\)+\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+...+\(\frac{1}{50^2}\)

A=1+\(\frac{1}{2^2}\)\(\frac{1}{3^2}\)+...+\(\frac{1}{50^2}\)

A<1+\(\frac{1}{1\cdot2}\)+\(\frac{1}{2\cdot3}\)+...+\(\frac{1}{49\cdot50}\)

A<1+1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+...+\(\frac{1}{49}\)-\(\frac{1}{50}\)

A<2-\(\frac{1}{50}\)<2

=>A<1(câu 1)

A= ^{\(\dfrac{1}{1^2}\)}