For the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, the major focus is $F, P_{1}, P_{2}, \cdots, P_{n}$ are $n$ 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 this
15. Let $O$ be the origin, $\angle P_{1} F O=\alpha$, then because $\angle P_{1} F P_{2}=\angle P_{2} F P_{3}=\cdots=\angle P_{\vee} F P_{1}=\frac{2 \pi}{n}$, so $\angle P, F O=\alpha-(i-1) \cdot \frac{2 \pi}{n}(1 \leqslant i \leqslant n)$.
Let the distance from $P_{1}$ to the directrix be $d_{2}$, then $d .-P_{1} F \cos \left(\alpha-(i-1) \cdot \frac{2 \pi}{n}\right)=\frac{a^{2}}{c}-c$.
By the properties of the ellipse, $P_{\mathrm{r}} F=\frac{c}{d} \cdot d_{i}$, so $d,\left(1-\frac{c}{a} \cos \left(\alpha-(i-1) \cdot \frac{2 \pi}{n}\right)\right)=\frac{a^{2}-c^{2}}{c}=\frac{b^{2}}{c}$,
$$
\frac{1}{d_{1}}=\frac{c\left(1-\frac{i}{a} \cos \left(\alpha-(i-1) \frac{2 \pi}{n}\right)\right)}{b^{2}}, \sum_{1}^{n} \frac{1}{d_{1}}=\frac{n c}{b^{2}}-\frac{c^{2}}{a b^{2}} \sum_{i=1}^{n} \cos \left(\alpha-(i-1) \frac{2 \pi}{n}\right) \text {. }
$$

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

we have $\sum_{i=1}^{\pi} \frac{1}{d_{1}}=\frac{n c}{b^{2}}$.