Ta thấy:
$\frac{1}{2^{2}} < \frac{1}{1.2}$
$\frac{1}{3^{2}} < \frac{1}{2.3}$
.....
$\frac{1}{100^{2}} < \frac{1}{99.100}$
=> $P < \frac{1}{1.2} + \frac{1}{2.3} + ....... + \frac{1}{99.100}$
$P < 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + .......+ \frac{1}{99} - \frac{1}{100}$
$P< 1-\frac{1}{100}<1$ (đpcm)