Answer
Diverges and $\lim\limits_{n \to \infty} a_n=\dfrac{1}{\sqrt [3] {e}}$
Work Step by Step
Consider $a_n=(1-\dfrac{1}{3n})^n$
Now, $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} (1-\dfrac{1}{3n})^n$
Thus, we have $=\lim\limits_{n \to \infty} (1+\dfrac{(-1/3)}{n})^n$
So, $\lim\limits_{n \to \infty} a_n=e^{-1/3}=\dfrac{1}{\sqrt [3] {e}}$
We need $\lim\limits_{n \to \infty} a_n =0$ for the series to converge, but we can see that $\lim\limits_{n \to \infty} a_n \ne 0$
Hence, The Series Diverges and $\lim\limits_{n \to \infty} a_n=\dfrac{1}{\sqrt [3] {e}}$
