Chứng minh 1/2^2 + 1/3^2 + … + 1/100^2 < 1

15000 câu hỏi ôn tập Toán (Phần 106)

Đề bài. Chứng minh $\frac{1}{2^2}+\frac{1}{3^2}+\mathrm{\dots }+\frac{1}{{100}^2}<1$.

Lời giải:

Ta có: $\frac{1}{2^2}<\frac{1}{1.2}$

$\frac{1}{3^2}<\frac{1}{2.3}$

….

$\frac{1}{{100}^2}<\frac{1}{99.100}$

Suy ra: $\frac{1}{2^2}+\frac{1}{3^2}+\mathrm{\dots }+\frac{1}{{100}^2}<\frac{1}{1.2}+\frac{1}{2.3}+\mathrm{\dots }+\frac{1}{99.100}$

$\frac{1}{2^2}+\frac{1}{3^2}+\mathrm{\dots }+\frac{1}{{100}^2}<1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\mathrm{\dots }+\frac{1}{99}-\frac{1}{100}$

$\frac{1}{2^2}+\frac{1}{3^2}+\mathrm{\dots }+\frac{1}{{100}^2}<1-\frac{1}{100}$

$\frac{1}{2^2}+\frac{1}{3^2}+\mathrm{\dots }+\frac{1}{{100}^2}<\frac{99}{100}<\frac{100}{100}=1$

Vậy $\frac{1}{2^2}+\frac{1}{3^2}+\mathrm{\dots }+\frac{1}{{100}^2}<1$.