Ta có: \(\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)

\(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)

...

\(\frac{1}{10^2}<\frac{1}{9\cdot10}=\frac19-\frac{1}{10}\)

Do đó: \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{10^2}<1-\frac12+\frac12-\frac13+\cdots+\frac19-\frac{1}{10}\)

=>\(A<1-\frac{1}{10}\)

=>A<1

Ta có: \(\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)

\(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)

...

\(\frac{1}{n^2}<\frac{1}{\left(n-1\right)\cdot n}=\frac{1}{n-1}-\frac{1}{n}\)

Do đó: \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}<1-\frac12+\frac12-\frac13+\cdots+\frac{1}{n-1}-\frac{1}{n}\)

=>\(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}<1-\frac{1}{n}<1\)

=>\(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}<1+1=2\)

=>A<2

Ta có: \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}>0\)

=>\(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}>1\)

=>A>1

Do đó: 1<A<2

=>A không là số tự nhiên

  1 2 2   < 1 1.2 ; 1 3 2 < 1 2.3 ; 1 4 2 < 1 3.4 ; ... ; 1 10 2 < 1 9.10

⇒ 1 2 2   + 1 3 2 + 1 4 2 + 1 10 2 < 1 1.2 + 1 2.3 + 1 3.4 + ... + 1 9.10 < 1.

1 2 2 + 1 3 2 + 1 4 2 + ... + 1 9 2 > 1 2.3 + 1 3.4 + 1 4.5 + ... + 1 9.10 = 2 5

1 2 2 + 1 3 2 + 1 4 2 + ... + 1 9 2 < 1 1.2 + 1 2.3 + 1 3.4 + 1 8.9 = 8 9

a ) 1 2.3 + 1 3.4 + ... + 1 6.7 = 1 2 − 1 7 < 1 2 .

b ) 4 1.5 + 4 5.9 + 4 9.13 + 4 13.17 + 4 17.21 = 1 − 1 21 < 1. c ) T a   c ó     1 2 2   < 1 1.2 ; 1 3 2 < 1 2.3 ; 1 4 2 < 1 3.4 ; ... ; 1 10 2 < 1 9.10 . D o   đ ó , 1 2 2   + 1 3 2 + 1 4 2 + 1 10 2 < 1 1.2 + 1 2.3 + 1 3.4 + ... + 1 9.10 < 1.

a ) 1 3.4 + 1 4.5 + ... + 1 19.20 = 1 3 − 1 20 = 17 60 < 1 2

b ) 3 1.4 + 3 4.7 + 3 7.10 + ... + 3 97.100 = 1 − 1 100 < 1

c ) T a   c ó   : 1 2 2 + 1 3 2 + 1 4 2 + ... + 1 9 2 > 1 2.3 + 1 3.4 + 1 4.5 + ... + 1 9.10 = 2 5

1 2 2 + 1 3 2 + 1 4 2 + ... + 1 9 2 < 1 1.2 + 1 2.3 + 1 3.4 + 1 8.9 = 8 9

a ) 1 1.2 + 1 2.3 + ... + 1 9.10 = 9 10 < 1.

b ) 3 2.5 + 3 5.8 + 3 8.11 + 3 11.14 + 3 14.17 + 3 17.20 = 1 2 − 1 20 < 1 2 .

c ) T a   c ó     1 2 2   < 1 1.2 ; 1 3 2 < 1 2.3 ; 1 4 2 < 1 3.4 ; ... ; 1 25 2 < 1 24.25 .

D o   đ ó , 1 2 2   + 1 3 2 + 1 4 2 + 1 25 2 < 1 1.2 + 1 2.3 + 1 3.4 + ... + 1 24.25 < 1.