`MATH`

ta có: $$\frac{1}{n^2} < \frac{1}{(n-1)n}$$ với mọi $n \geq 2$$

khi đó:

$$A = \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \dots + \frac{1}{100^2}$$

$$A < \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \dots + \frac{1}{99 \cdot 100}$$

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

$$A < 1 - \frac{1}{100}$$

$$A < \frac{99}{100}$$

Vì $\frac{99}{100} < 1$, mà $1 < 4$ (tức $1 \cdot 4$).

Vậy $A < 4$ (Đpcm).