Answer
The limit does not exist.
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 $a_n=\dfrac{(-1)^n \ n}{n+1}$ and $f(x)=\dfrac{(-1)^x \ x}{x+1}$
Next, $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} \dfrac{(-1)^x \ x}{x+1}=\lim\limits_{n \to \infty} \dfrac{(-1)^x}{(1+\dfrac{1}{x})}$
a) When $n$ is odd.
$\lim\limits_{n \to \infty} a_n=\lim\limits_{x \to \infty} \dfrac{-1}{(1+\dfrac{1}{x})}=\dfrac{-1}{1+0}=-1$
b) When $n$ is even.
$\lim\limits_{n \to \infty} a_n=\lim\limits_{x \to \infty} \dfrac{1}{(1+\dfrac{1}{x})}=\dfrac{1}{1+0}=1$
Thus. we see that the limit of $a_n$ is different in both cases, so the limit does not exist.
