4. Let $P$ be any point on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, with foci at $F_{1}(-c, 0), F_{2}(c, 0)$. $P F_{1}$ and $P F_{2}$ intersect the ellipse at points $A$ and $B$, respectively. If $a^{2}$, $b^{2}$, and $c^{2}$ form an arithmetic sequence, then $\frac{\left|P F_{1}\right|}{\left|A F_{1}\right|}+\frac{\left|P F_{2}\right|}{\left|B F_{2}\right|}=$ $\qquad$.
4.4.

As shown in Figure 1.
By the definition of an ellipse, the similarity of triangles, and the properties of ratios, we have
$$
\begin{array}{l} 
e=\frac{\left|P F_{1}\right|}{|P Q|}=\frac{\left|A F_{1}\right|}{|A C|} \Rightarrow \frac{\left|P F_{1}\right|}{\left|A F_{1}\right|}=\frac{|P Q|}{|A C|} \\
\Rightarrow \frac{|P A|}{\left|A F_{1}\right|}=\frac{|P Q|+|A C|}{|A C|} . \\
\text { Also, } \frac{|P A|}{\left|A F_{1}\right|}=\frac{|P D|}{\left|E F_{1}\right|}=\frac{|P Q|-|A C|}{\left|M F_{1}\right|-|A C|} \\
=\frac{2|P Q|}{\left|M F_{1}\right|}=\frac{2\left|P F_{1}\right|}{e\left|M F_{1}\right|} \\
=\frac{2\left|P F_{1}\right|}{\frac{c}{a}\left(\frac{a^{2}}{c}-c\right)}=\frac{2 a\left|P F_{1}\right|}{b^{2}}, \\
\text { Therefore, } \frac{\left|P F_{1}\right|}{\left|A F_{1}\right|}=\frac{|P A|}{\left|A F_{1}\right|}-1=\frac{2 a\left|P F_{1}\right|}{b^{2}}-1 .
\end{array}
$$

Similarly, $\frac{\left|P F_{2}\right|}{\left|B F_{2}\right|}=\frac{2 a\left|P F_{2}\right|}{b^{2}}-1$.
Since $a^{2}+c^{2}=2 b^{2}$, we have
$$
\begin{array}{l}
\frac{\left|P F_{1}\right|}{\left|A F_{1}\right|}+\frac{\left|P F_{2}\right|}{\left|B F_{2}\right|} \\
=\frac{2 a\left|P F_{1}\right|}{b^{2}}-1+\frac{2 a\left|P F_{2}\right|}{b^{2}}-1=\frac{4 a^{2}}{b^{2}}-2 \\
=\frac{2\left(2 a^{2}-b^{2}\right)}{b^{2}}=\frac{2\left(a^{2}+c^{2}\right)}{b^{2}}=4 .
\end{array}
$$