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

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.
[Solution] From the given information,
$$\begin{aligned}
T_{n}= & S_{1}+S_{2}+S_{3}+\cdots+S_{n} \\
= & 1+\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{2}+\frac{1}{3}\right)+\cdots+\left(1+\frac{1}{2}+\frac{1}{3}+\right. \\
& \left.\cdots+\frac{1}{n}\right) \\
= & n+\frac{1}{2}(n-1)+\frac{1}{3}(n-2)+\cdots+\frac{1}{n}[n-(n-1)] \\
= & n\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)-\left[\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)\right. \\
& \left.+\cdots+\left(1-\frac{1}{n}\right)\right] \\
= & n S_{n}-\left[(n-1)-\left(S_{n}-1\right)\right] \\
= & (n+1) S_{n}-n \\
= & (n+1) S_{n+1}-(n+1) \\
\therefore \quad T_{n}= & (n+1) S_{n}-n=(n+1) S_{n+1}-(n+1)
\end{aligned}$$

Let $n=1988$, then we get
$$T_{1988}=1989 S_{1989}-1989$$

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

Let $n=1988$, we get
$$U_{1988}=1990 S_{1989}-3978$$

Therefore, $c=1990, d=3978$.