Answer
$0$
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, we have $a_n=\dfrac{(-1)^{n+1} \ n^2}{2n^3+n}$ and $f(x)=\dfrac{(-1)^{x+1} \ x^2}{2x^3+x}$
Next, $\lim\limits_{n \to \infty} a_n=\lim\limits_{x \to \infty} \dfrac{(-1)^{x+1} \ x^2}{2x^3+x}=\lim\limits_{x \to \infty} \dfrac{(-1)^{x+1} (1/x)}{(2+\dfrac{1}{x^2})}$
a) When $n$ is odd and $n+1$ is even.
$\lim\limits_{n \to \infty} a_n=\lim\limits_{x \to \infty} \dfrac{-1 (1/x)}{(2+\dfrac{1}{x^2})}=\dfrac{-1(0)}{2+0}=0$
b) When $n$ is even and $n+1$ is odd.
$\lim\limits_{n \to \infty} a_n=\lim\limits_{x \to \infty} \dfrac{1 (1/x)}{(2+\dfrac{1}{x^2})}=\dfrac{(1)(0)}{2+0}=0$
Thus,$\lim\limits_{n \to \infty} a_n=0$