Let $P_{n}=(1+1)\left(1+\frac{1}{4}\right)\left(1+\frac{1}{7}\right) \cdots$ $\left(1+\frac{1}{3 n-2}\right)$. Find the greatest integer part of $P_{2000}$.
$$
\begin{array}{l}
\text{First, prove the following inequality:} \\
\sqrt[3]{\frac{7 n+1}{7 n-6}}1) .
\end{array}
$$

The inequality (1) is equivalent to
$$
\begin{array}{l}
\frac{7 n+1}{7 n-6}0 \\
\Leftrightarrow 27(n-1)^{2}+13(n-1)>0 .
\end{array}
$$

The last inequality holds for $n>1$, thus inequality (1) holds.
Similarly, inequality (2) ( (2) $\Leftrightarrow n>1$ ) can also be proven to hold.
Next, find $\left[P_{2000}\right]$.
From inequalities (1) and (2), we get
$$
\begin{array}{l}
P_{2000}>(1+1) \times \sqrt[3]{\frac{15}{8}} \times \sqrt[3]{\frac{22}{15}} \times \ldots \times \\
\sqrt[3]{\frac{72000+1}{7 \times 2000-6}} \\
=\sqrt[3]{14001} \doteq 24.1, \\
P_{2000}<(1+1) \times \sqrt[3]{\frac{2}{1}} \times \sqrt[3]{\frac{3}{2}} \times \ldots \times \sqrt[3]{2000} \\
=\sqrt[3]{16000} \doteq 25.2 .
\end{array}
$$

Therefore, $\left[P_{2000}\right]=25$.