Question

A = \(\frac{1}{2^2}\) + \(\frac{1}{3^2}\) + \(\frac{1}{4^2}\) + ....+ \(\frac{1}{100^2}\) < 1

Answer 1

Đặt \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

Ta có:

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

Mà \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}< 1\left(2\right)\)

Từ (1) và (2) suy ra \(A< B< 1\Rightarrow A< 1\)

Đpcm

Answer 2

Ta có:
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

Mà \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}< 1\)

\(\Rightarrow A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}< 1\)

hay \(A< 1\left(đpcm\right)\)