10. (20 points) Given the sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy:
$$
a_{1}=a_{2}=2019, b_{1}=b_{2}=2017 \text {, }
$$

and for all positive integers $n \geqslant 2$, we have
$$
\begin{array}{l}
a_{n+1}=\left(a_{n-1}-1\right)\left(b_{n}+1\right), \\
b_{n+1}=a_{n} b_{n-1}-1 . \\
\text { Find } \sum_{n=1}^{\infty} b_{n}\left(\frac{1}{a_{n+1}}-\frac{1}{a_{n+3}}\right) \text { . }
\end{array}
$$
10. First, define $a_{-1}=a_{0}=b_{-1}=b_{0}=0$.

Using mathematical induction, it can be proven that for all positive integers $k$, we have
$$
\begin{array}{l}
a_{k}>a_{k-2}+1, b_{k}>b_{k-2}+1, \\
a_{k}>2000, b_{k}>2000 .
\end{array}
$$

Thus, $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ are positive, increasing sequences that tend to positive infinity.
For $n \in \mathbf{Z}_{+}$, we have
$$
\begin{array}{l}
a_{n+3}=\left(a_{n+1}-1\right)\left(b_{n+2}+1\right) \\
=\left(a_{n+1}-1\right) a_{n+1} b_{n} . \\
\text { Then } \frac{b_{n}}{a_{n+3}}=\frac{1}{a_{n+1}\left(a_{n+1}-1\right)} \\
=\frac{1}{a_{n+1}-1}-\frac{1}{a_{n+1}} . \\
\text { Also, when } n \geqslant 2 \text {, we have } \\
\frac{b_{n}+1}{a_{n+1}}=\frac{1}{a_{n-1}-1} . \\
\text { Therefore, } \sum_{n=1}^{\infty} b_{n}\left(\frac{1}{a_{n+1}}-\frac{1}{a_{n+3}}\right)=\sum_{n=1}^{\infty}\left(\frac{b_{n}}{a_{n+1}}-\frac{b_{n}}{a_{n+3}}\right) \\
=\sum_{n=1}^{\infty}\left(\frac{b_{n}}{a_{n+1}}-\frac{1}{a_{n+1}-1}+\frac{1}{a_{n+1}}\right) \\
\quad=\sum_{n=1}^{\infty}\left(\frac{b_{n}+1}{a_{n+1}}-\frac{1}{a_{n+1}-1}\right) \\
\quad=\frac{b_{1}+1}{a_{2}}-\frac{1}{a_{2}-1}+\sum_{n=2}^{\infty}\left(\frac{1}{a_{n-1}-1}-\frac{1}{a_{n+1}-1}\right) \\
=\frac{b_{1}+1}{a_{2}}-\frac{1}{a_{2}-1}+\left(\frac{1}{a_{1}-1}+\frac{1}{a_{2}-1}\right) \\
=\frac{b_{1}+1}{a_{2}}+\frac{1}{a_{1}-1} \\
=\frac{4074343}{4074} .
\end{array}
$$