4. Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$

with points $A$ and $B$ 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) $\left(\frac{1}{3}, 1\right)$
4. C.

As shown in Figure 2.
From $\overrightarrow{F_{1} A}=3 \overrightarrow{F_{2} B}$, we get
$F_{2} B \parallel F_{1} A, \left|F_{2} B\right|=\frac{1}{3}\left|F_{1} A\right|$.
Thus, the line $A B$ intersects the $x$-axis at point $P(2 c, 0)$.
Let $M$ be a moving point on the ellipse, and the point that divides $P M$ into a ratio of $1: 2$ forms the curve $C$. Then the intersection points of curve $C$ and the ellipse are the points $B$.

Clearly, if point $P$ is inside the ellipse, then curve $C$ is also inside the ellipse and has no common points with the ellipse; whereas if point $P$ is outside the ellipse, then curve $C$ intersects the ellipse.

Therefore, the necessary and sufficient condition for the existence of two points $A$ and $B$ on the ellipse such that $\overrightarrow{F_{1} A}=3 \overrightarrow{F_{2} B}$ is that point $P$ is outside the ellipse.
Hence $2 c > a \Rightarrow e=\frac{c}{a}>\frac{1}{2}$.