Five, given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ with the left focus at $F, P_{1}, P_{2}, \cdots, P_{n}$ being $n$ points arranged in a clockwise order on the ellipse, and $\angle P_{1} F P_{2}=\angle P_{2} F P_{3}=\angle P_{3} F P_{4}=\cdots$ $=\angle P_{n} F P_{1}$,
find the sum of the reciprocals of the distances from these $n$ points to the left directrix $x=-\frac{a^{2}}{c}\left(c=\sqrt{a^{2}-b^{2}}\right)$ (expressed in terms of $a, b, c, n$).
Let $O$ be the origin, $\angle P_{1} F O=\alpha$,
then because $\angle P_{1} F P_{2}=\angle P_{2} F P_{3}=\cdots=\angle P_{N} F P_{1}=\frac{2 \pi}{n}$,
so $\angle P_{i} F O=\alpha-(i-1) \cdot \frac{2 \pi}{n}(1 \leqslant i \leqslant n)$
Let the distance from $P_{i}$ to the left directrix be $d_{i}$, then
$$
d_{i}-P_{i} F \cos \left(\alpha-(i-1) \cdot \frac{2 \pi}{n}\right)=\frac{a^{3}}{c}-c
$$

By the property of the ellipse,
$$
P_{i} F=\frac{c}{d} \cdot d_{i}
$$

So
$$
\begin{array}{c}
d_{i}\left(1-\frac{c}{a} \cos \left(\alpha-(i-1) \cdot \frac{2 \pi}{n}\right)\right)=\frac{a^{2}-c^{2}}{c}=\frac{b^{2}}{c}, \\
\frac{1}{d_{i}}=\frac{c\left(1-\frac{c}{a} \cos \left(\alpha-(i-1) \frac{2 \pi}{n}\right)\right)}{b^{2}}, \\
\sum \frac{1}{d_{i}}=\frac{n c}{b^{2}}-\frac{c^{2}}{a b^{2}} \sum_{i=1}^{n} \cos \left(\alpha-(i-1) \frac{2 \pi}{n}\right)
\end{array}
$$

Because
$$
\begin{aligned}
& \sum_{i=1}^{n} \cos \left(\alpha-(i-1) \frac{2 \pi}{n}\right) \\
= & \frac{1}{2 \sin \frac{\pi}{n}} \sum\left(\sin \left(\alpha-(i-1) \frac{2 \pi}{n}+\frac{\pi}{n}\right)-\sin \left(\alpha-(i-1) \frac{2 \pi}{n}-\frac{\pi}{n}\right)\right) \\
= & \frac{1}{2 \sin \frac{\pi}{n}}\left(\sin \left(\alpha-(n-1) \frac{2 \pi}{n}+\frac{\pi}{n}\right)-\sin \left(\alpha-\frac{\pi}{n}\right)\right)=0
\end{aligned}
$$

So $\sum \frac{1}{d_{i}}=\frac{n c}{b^{2}}$.