Problem 278

Solution:
(1) $S_1=1- \frac{1}{2}= \frac{1}{2}$, $S_2=1- \frac{1}{2}+ \frac{1}{3}- \frac{1}{4}= \frac{7}{12}$
$T_1= \frac{1}{1+1}= \frac{1}{2}$, $T_2= \frac{1}{2+1}+ \frac{1}{2+2}= \frac{7}{12}$
(2) Conjecture: $S_n=T_n$ ($n\in\mathbb{N}^*$), that is:
$1- \frac{1}{2}+ \frac{1}{3}- \frac{1}{4}+\cdots + \frac{1}{2n-1}- \frac{1}{2n}= \frac{1}{n+1}+ \frac{1}{n+2}+ \frac{1}{n+3}+\cdots + \frac{1}{2n}$.
Now, let's prove it using mathematical induction:
(i) When $n=1$, it has been proven that $S_1=T_1$.
(ii) Assume when $n=k$ ($k\geqslant 1, k\in\mathbb{N}^*$), $S_k=T_k$,
that is: $1- \frac{1}{2}+ \frac{1}{3}- \frac{1}{4}+\cdots + \frac{1}{2k-1}- \frac{1}{2k}= \frac{1}{k+1}+ \frac{1}{k+2}+ \frac{1}{k+3}+\cdots + \frac{1}{2k}$.
Then: $S_{k+1}=S_k+ \frac{1}{2k+1}- \frac{1}{2(k+1)}=T_k+ \frac{1}{2k+1}- \frac{1}{2(k+1)}$
$= \frac{1}{k+1}+ \frac{1}{k+2}+ \frac{1}{k+3}+\cdots + \frac{1}{2k}+ \frac{1}{2k+1}- \frac{1}{2(k+1)}$
$= \frac{1}{k+2}+ \frac{1}{k+3}+\cdots + \frac{1}{2k+1}+ (\frac{1}{k+1}- \frac{1}{2(k+1)})$
$= \frac{1}{(k+1)+1}+ \frac{1}{(k+1)+2}+\cdots + \frac{1}{2k+1}+ \frac{1}{2(k+1)}=T_{k+1}$,
From (i) and (ii), it is known that for any $n\in\mathbb{N}^*$, $S_n=T_n$ holds.
Therefore, the final answer is $\boxed{S_n=T_n}$.