Question

Chứng Minh : \(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< 2\)

Answer 1

1/1^2 + 1/2^2 +1/3^2 +.....+1/100^2 nhỏ hơn 1+1/1×2+1/2×3+....+1/99.100= 1+1-1/2+1/2-1/3 +...1...1/99-1/100 suy ra 2 -1/100 nhỏ hơn 2 suy ra đpcm

Answer 2

\(\frac{1}{1^2}=1\)

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

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

...

\(\frac{1}{100^2}< \frac{1}{99.100}\)

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

Vậy \(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< 2\)