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Thấy $\frac{1}{k}>\frac{1}{k+1}(k\in Z, k\geq 1)\Rightarrow \frac{1}{k}-\frac{1}{k+1}>0$
$P=\frac{1}{2}-\left ( \frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2005}-\frac{1}{2006} \right )$
$P<\frac{1}{2}-\left ( \frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6} \right)=\frac{1}{2}-\frac{7}{60}<\frac{1}{2}-\frac{1}{10}=\frac{2}{5}$
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