Eight, Buzhi Line
$1_{1:} \mathrm{y}=\mathrm{kx}+1$ ( 0
$<\mathrm{k}<1$ ), $1_{2}$
$y=x$. Through the intersection point $P_{0}$ of $l_{1}$ and the $y$-axis, draw a line parallel to the $x$-axis to intersect $1_{2}$ at point
$Q_{13}$ Draw a line parallel to the $y$-axis through $Q_{1}$ to intersect $I_{1}$ at $P_{1}$,
Draw a line parallel to the $x$-axis through $P_{1}$ to intersect $1_{2}$ at $Q_{23}$ Draw a line parallel to the $y$-axis through $Q_{2}$ to intersect $1_{1}$ at $P_{2,}, \cdots$ Continue this process infinitely, obtaining points $Q_{3} 、 P_{3} 、 Q_{4} 、 P_{4}, \ldots$
(1) Let the line segment $p_{n}-1 Q_{3}=a_{n}$, find $a_{1} 、 a_{2}, a_{3}$, and
(2) Find the areas of $\triangle P_{0} Q_{1} P_{1}, \triangle P_{1} Q_{2} P_{2}, \triangle P_{2} Q_{3} P_{3}, \cdots$.
(1) $\because P_{0}(0,1), Q_{1}(1,1)$,
$$
\begin{aligned}
\therefore & a_{1}=P_{0} Q_{1}=1, \\
& \because P_{1}(1, k+1), Q_{2}(k+1, k+1), \\
& \therefore a_{2}=P_{1} Q_{2}=k . \\
& \because P_{2}\left(k+1, k^{2}+k+1\right), \\
& Q_{3}\left(k^{2}+k+1, k^{2}+k+1\right), \\
& \therefore a_{3}=P_{2} Q_{3}=k^{2}, \cdots, \\
\therefore a_{s}= & P_{s-1} Q_{s}=k^{s-1} .
\end{aligned}
$$
(2) $\because P_{s-1} Q_{s}=k^{s-1}, Q_{s} P_{s}=k^{s}$,
$$
\begin{array}{l}
\therefore \Delta P_{s-1} Q_{s} P_{s} \text { area }=\frac{1}{2} P_{s-1} Q_{s} \cdot P_{s} Q_{s} \\
\quad=\frac{1}{2} k^{s-1} \cdot k^{s}=\frac{1}{2} k^{2s-1}=\frac{1}{2 k}\left(k^{s}\right)^{2} . \\
\because 0<k<1, \therefore 0<k^{2}<1 .
\end{array}
$$
$\therefore \triangle P_{0} Q_{1} P_{1}, \triangle P_{1} Q_{2} P_{2}, \cdots$ form an infinite decreasing geometric series, let the sum of all terms be $S$, then
$$
S=\frac{1}{2 k} \cdot \frac{k^{2}}{1-k^{2}}=\frac{k}{2\left(1-k^{2}\right)} .
$$