4. Given $F_{1}, F_{2}$ are the left and right foci of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, and there exist two points $A, B$ on the ellipse such that $\overrightarrow{F_{1} A}=3 \overrightarrow{F_{2} B}$, then the range of the eccentricity of the ellipse is
A. $\left(0, \frac{1}{2}\right)$
B. $\left(0, \frac{1}{3}\right)$
C. $\left(\frac{1}{2}, 1\right)$
D. $\left(\frac{1}{3}, 1\right)$
Extend $A F_{1}$ to intersect the ellipse at point $C$. By symmetry, we know $\left|F_{1} C\right|=\left|F_{2} B\right|$. When $\left|F_{1} A\right|$ is maximized, $\left|F_{1} C\right|$ is minimized. Thus, $a+c>3(a-c) \Rightarrow 2 c>a \Rightarrow e \in\left(\frac{1}{2}, 1\right)$, so the answer is $C$.