Solution:
(1) When $n=1$, $f(1)=1$, $g(1)=2( \sqrt {2}-1)$, $f(1) > g(1)$,
When $n=2$, $f(2)=1+ \frac {1}{ \sqrt {2}}$, $g(2)=2( \sqrt {3}-1)$, $f(2) > g(2)$,
When $n=3$, $f(3)=1+ \frac {1}{ \sqrt {2}}+ \frac {1}{ \sqrt {3}}$, $g(3)=2$, $f(3) > g(3)$.
(2) Conjecture: $f(n) > g(n) (n\in\mathbb{N}^{*})$, i.e., $1+ \frac {1}{ \sqrt {2}}+ \frac {1}{ \sqrt {3}}+\ldots+ \frac {1}{ \sqrt {n}} > 2( \sqrt {n+1}-1) (n\in\mathbb{N}^{*})$.
Now, let's prove it by mathematical induction:
① When $n=1$, it has been proven above.
② Assume the conjecture holds for $n=k$, i.e., $1+ \frac {1}{ \sqrt {2}}+ \frac {1}{ \sqrt {3}}+\ldots+ \frac {1}{ \sqrt {k}} > 2( \sqrt {k+1}-1)$
Then when $n=k+1$, $f(k+1)=1+ \frac {1}{ \sqrt {2}}+ \frac {1}{ \sqrt {3}}+\ldots+ \frac {1}{ \sqrt {k}}+ \frac {1}{ \sqrt {k+1}} > 2( \sqrt {k+1}-1)+ \frac {1}{ \sqrt {k+1}}=2 \sqrt {k+1}+ \frac {1}{ \sqrt {k+1}}-2$;
And $g(k+1)=2( \sqrt {k+2}-1)=2 \sqrt {k+2}-2$, next we need to prove: $2 \sqrt {k+1}+ \frac {1}{ \sqrt {k+1}} > 2 \sqrt {k+2}$
It suffices to prove: $2(k+1)+1=2k+3 > 2 \sqrt {(k+2)(k+1)}$, it requires to prove: $(2k+3)^{2} > 4(k+2)(k+1)$,
i.e., $4k^{2}+12k+9 > 4k^{2}+12k+8$, this inequality obviously holds. Therefore, the conjecture also holds for $n=k+1$.
In summary, for $n\in\mathbb{N}^{*}$, the conjecture holds,
i.e., $1+ \frac {1}{ \sqrt {2}}+ \frac {1}{ \sqrt {3}}+\ldots+ \frac {1}{ \sqrt {n}} > 2( \sqrt {n+1}-1) (n\in\mathbb{N}^{*})$ is true. $\boxed{\text{True}}$