Answer
Converges
Work Step by Step
Let us consider that $f(x)=\dfrac{x}{x^2-3}$ for $x \gt 0$
On differnitating, we get $f'(x)=-\dfrac{x^2+3}{(x^2-3)^2}$
It has been clearly determined that $f'(x) \lt 0$ for all $x \geq 1$. This implies that the function $f(x)$ shows a monotonically decreasing sequence for $x \geq 1$
and $\lim\limits_{x \to \infty} f(x)=0$
Next, $f(x) \gt 0$ for all $x \geq 1$, So, the given series is convergent by the Leibniz's Test or the Alternating Series Test.
