(1) When $n=1$, $f(1)=1=g(1)$;
When $n=2$, $f(2)= \frac{9}{8}$, $g(2)= \frac{11}{8}$, $\therefore f(2) < g(2)$;
When $n=3$, $f(3)= \frac{251}{216}$, $g(3)= \frac{312}{216}$, $\therefore f(3) < g(3)$.
(2) Based on (1), we conjecture that $f(n) \leqslant g(n)$. We will prove this using mathematical induction:
$\quad \quad$ $\bullet$ When $n=1$, $2$, $3$, the inequality holds.
$\quad \quad$ $\bullet$ Assume that the inequality holds when $n=k (k \in \mathbb{N}^*) (k \geqslant 3)$, i.e., $1+ \frac{1}{2^{2}}+ \frac{1}{3^{2}}+…+ \frac{1}{k^{2}} < \frac{1}{2}(3- \frac{1}{k^{2}})$.
Then, when $n=k+1$, we have $f(k+1)=f(k)+ \frac{1}{(k+1)^{2}} < \frac{1}{2}(3- \frac{1}{k^{2}})+ \frac{1}{(k+1)^{2}}$.
$\quad \quad$ $\quad \quad$ Since $\frac{1}{2(k+1)^{2}}- \frac{1}{2k^{2}}+ \frac{1}{(k+1)^{2}}= \frac{-3k-1}{2(k+1)^{2}k^{2}} < 0$, $\therefore - \frac{1}{2k^{2}}+ \frac{1}{(k+1)^{2}} < - \frac{1}{2(k+1)^{2}}$,
$\quad \quad$ $\quad \quad$ $\therefore f(k+1) < \frac{3}{2}- \frac{1}{2(k+1)^{2}} = \boxed{g(k+1)}$, i.e., the inequality holds when $n=k+1$.
Thus, by mathematical induction, we conclude that $f(n) \leqslant g(n)$ for all $n \in \mathbb{N}^*$.