Question

[Đặng Tú Phương]

Cho tổng A=\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...\frac{1}{99}+\frac{1}{100}\)

Chứng tỏ rằng A > 1

Answer 1

[Dũng Lê Trí]

\(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\)

\(=\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\right)>\frac{1}{10}+\left(\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\right)\)

\(=\frac{1}{10}+\frac{90}{100}>1\)

\(A>1\left(đpcm\right)\)

Answer 2

[Thắng Hoàng]

a>1(đpcm)