Question

So sánh A = 1 + 1/2² + 1/3² + 1/4² + ... + 1/100² với 2

Answer 1

\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

mà \(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}=2-\frac{1}{100}< 2\)

Vậy \(A< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}< 2\)

=>A<2(đpcm)

Answer 2

Ta có: A = 1 + 1/2²+1/3²+1/4^2+...+1/100^2 < 1+1/1.2 +1/2.3+1/3.4+...+1/99.100

     => A < 1+(1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100)

          A < 1+(1-1/100)

          A < 1+99/100

Vì 1+99/100  < 2 nên A < 2

Answer 3

vì 1/2^2+.....+1/100^2 đã bé hơn 1 nên tất nhiên là phải bé hơn 2