Example 5 Find the successive convergents of the finite simple continued fraction $\langle 1,1,1,1,1,1,1,1,1,1\rangle$.
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Solution Of course, we can calculate one by one using the method in Example 1, but at this point, it is more convenient to use Theorem 2 to recursively calculate $P_{n}, Q_{n}$. According to formulas (38), (37), the following table can be listed, where $x_{n}=1,0 \leqslant n \leqslant 9$ and $P_{-2}=0, P_{-1}=1, Q_{-2}=1, Q_{-1}=0$,
$$P_{n}=P_{n-1}+P_{n-2}, \quad Q_{n}=Q_{n-1}+Q_{n-2}, \quad n \geqslant 0$$
\begin{tabular}{|c|cccccccccc|}
\hline$n$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline$x_{n}$ & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
$P_{n}$ & 1 & 2 & 3 & 5 & 8 & 13 & 21 & 34 & 55 & 89 \\
$Q_{n}$ & 1 & 1 & 2 & 3 & 5 & 8 & 13 & 21 & 34 & 55 \\
\hline
\end{tabular}

Therefore, the successive convergents $P_{n} / Q_{n}$ are $1 / 1,2 / 1,3 / 2,5 / 3,8 / 5,13 / 8$, $21 / 13,34 / 21,55 / 34,89 / 55$. It is evident that $P_{n}, Q_{n}$ are both Fibonacci sequences, and we have
$$P_{n}=Q_{n+1}, \quad n \geqslant 0$$
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