The answer is $\boxed{\frac{29}{53}}$.
The sequence is a geometric sequence with a common ratio of $\frac{2}{3}$. This means that each term is $\frac{2}{3}$ times the term before it.
The first term is $\frac{4}{9}$. The second term is $\frac{4}{9} \times \frac{2}{3} = \frac{9}{20}$. The third term is $\frac{9}{20} \times \frac{2}{3} = \frac{18}{30}$. The fourth term is $\frac{18}{30} \times \frac{2}{3} = \frac{29}{53}$.
Here is a table of the terms of the sequence:
Term | Value
——- | ——–
1 | $\frac{4}{9}$
2 | $\frac{9}{20}$
3 | $\frac{18}{30}$
4 | $\frac{29}{53}$
Here is a graph of the sequence:
[asy]
unitsize(1 cm);
real ticklen = 1;
real tickspace = 1;
real axisarrowsize = 0.14inch;
real tickdown = -0.5;
real tickdownlength = -0.12inch;
real tickdownbase = -0.25inch;
real wholetickdown = tickdown;
real wholetickdownlength = tickdownlength;
real wholetickdownbase = tickdownbase;
real axiswidth = 0.12inch;
real linetype = 1;
real linewidth = 0.7pt;
real dotsep = 3pt;
real dotsize = 3pt;
real g = 1.2;
real ticklength = 0.14inch;
real axismargin = 0.1inch;
real xmargin = 1.2inch;
real ymargin = 1.2inch;
draw((0,-1.2)–(0,1.2),Arrows(0.14inch));
draw((-1.2,0)–(1.2,0),Arrows(0.14inch));
label(“$x$”,(1.2,0),SE);
label(“$y$”,(0,1.2),NW);
real i;
for (i=-1; i<=1; ++i) {
draw((i,-0.1)–(i,0.1));
}
real j;
for (j=-1; j<=1; ++j) {
draw((0,j)–(0.1,j));
}
real t;
for (t=0; t<=1; t+=0.1) {
draw((t,-0.05)–(t,0.05));
}
real x;
real y;
x = 0;
y = 4/9;
draw((x,y)–(x+0.1,y),linetype,linewidth);
x = 0;
y = 9/20;
draw((x,y)–(x+0.1,y),linetype,linewidth);
x = 0;
y = 18/30;
draw((x,y)–(x+0.1,y),linetype,linewidth);
x = 0;
y = 29/53;
draw((x,y)–(x+0.1,y),linetype,linewidth);
label(“$\frac{4}{9}$”,(-0.1,4/9),SE);
label(“$\frac{9}{20}$”,(-0.1,9/20),SE);
label(“$\frac{18}{30}$”,(-0.1,18/30),SE);
label(“$\frac{29}{53}$”,(-0.1,29/53),SE);
[/asy]