Answer
$a_n=\frac{2^{n-1}}{3(n+2)}$
Work Step by Step
We can see that the subsequent terms' numerator gets multiplied by $2$ and $3$ is added to the denominator with $a_1=\frac{1}{9}$, so $a_n=\frac{2^{n-1}}{3(n+2)}$ for all $n$ natural numbers.