Answer
converges to $0$
Work Step by Step
Definition of a sequence defined by a function: Let us consider a function $f(x)$ whose limit $\lim\limits_{x \to \infty} f(x)$ exists then the sequence $a_n=f(n)$ will converge to the same limit.
That is, $\lim\limits_{n \to \infty} a_n=\lim\limits_{x \to \infty} f(x)$
Here, in this problem we have $a_n=\dfrac{2 \tan^{-1} n}{n^3+4}$ and $f(x)=\dfrac{2 \tan^{-1} x}{x^3+4}$
Next, $\lim\limits_{n \to \infty} a_n=\lim\limits_{x \to \infty} \dfrac{2 \tan^{-1} x}{x^3+4}$
or, $\lim\limits_{n \to \infty} a_n= \dfrac{2 \lim\limits_{x \to \infty} \tan^{-1} x}{\lim\limits_{x \to \infty}(x^3+4)}=\dfrac{2 \tan^{-1} (\infty)}{\infty}$
or, $\lim\limits_{n \to \infty} a_n= \dfrac{(2) (\dfrac{\pi}{2})}{\infty}$
Thus. $\lim\limits_{n \to \infty} a_n=0$;so the sequence converges to $0$ .
