$3 \cdot 25$ For each positive integer $n$, let
$$
\begin{array}{l}
S_{n}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}, \\
T_{n}=S_{1}+S_{2}+S_{3}+\cdots+S_{n}, \\
U_{n}=\frac{1}{2} T_{1}+\frac{1}{3} T_{2}+\frac{1}{4} T_{3}+\cdots+\frac{1}{n+1} T_{n} .
\end{array}
$$

Try to find integers $0<a, b, c, d<1000000$, such that
$$
\begin{array}{l}
T_{1988}=a S_{1989}-b, \\
U_{1988}=c S_{1989}-d .
\end{array}
$$

And prove your result.
\[ \text{Solution:} \]
\[
\begin{array}{l}
T_{n}= S_{1}+S_{2}+S_{3}+\cdots+S_{n} \\
=1+\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{2}+\frac{1}{3}\right)+\cdots+\left(1+\frac{1}{2}+\frac{1}{3}+\right. \\
\left.\cdots+\frac{1}{n}\right) \\
= n+\frac{1}{2}(n-1)+\frac{1}{3}(n-2)+\cdots+\frac{1}{n}[n-(n-1)] \\
= n\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)-\left[\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)\right. \\
\left.+\cdots+\left(1-\frac{1}{n}\right)\right] \\
= n S_{n}-\left[(n-1)-\left(S_{n}-1\right)\right] \\
=(n+1) S_{n}-n \\
=(n+1) S_{n+1}-(n+1) \\
\therefore \quad T_{n}=(n+1) S_{n}-n=(n+1) S_{n+1}-(n+1) .
\end{array}
\]

Let \( n=1988 \), then we get
\[
T_{1988}=1989 S_{1989}-1989.
\]

Therefore, \( a=1989, b=1989 \).
\[
\begin{aligned}
\text{Also, } U_{n} & =\sum_{i=1}^{n} \frac{1}{i+1} T_{i}=\sum_{i=2}^{n+1} \frac{1}{i} T_{i-1} \\
& =\sum_{i=2}^{n+1} \frac{1}{i}\left(i S_{i}-i\right)=\sum_{i=2}^{n+1}\left(S_{i}-1\right) \\
& =\sum_{i=1}^{n+1}\left(S_{i}-1\right)=T_{n+1}-(n+1) \\
& =(n+2) S_{n+1}-(n+1)-(n+1) \\
& =(n+2) S_{n+1}-2(n+1), \\
\therefore \quad U_{n} & =(n+2) S_{n+1}-2(n+1) .
\end{aligned}
\]

Let \( n=1988 \), we get
\[
U_{1988}=1990 S_{1989}-3978.
\]