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

Try to find integers $0<a, b, c, d<1000000$, such that $T_{1988}=a \cdot S_{1989}-b, \quad U_{1988}=c \cdot S_{1989}$ $-d$. And prove your result.
\[
\begin{array}{l}
\text { 1. Solve } T_{\mathrm{n}}=\sum_{k=1}^{n} \sum_{1=1}^{k} \frac{1}{i}=\sum_{1=1}^{n} \sum_{k=1}^{n} \frac{1}{i} \\
=\sum_{1=1}^{n} \frac{n+1-i}{i}=(n+1) S_{0}-n \\
\quad=(n+1) S_{n+1}-(n+1), \\
\begin{aligned} 
\therefore \quad a=1989=b . \\ 
U_{0}=\sum_{1=1}^{n} \frac{1}{i+1} T_{1}=\sum_{i=2}^{n+1}\left(S_{1}-1\right) \\ 
=\sum_{i=1}^{\mathrm{a}+1}\left(S_{1}-1\right)=(n+2) S_{n}+1-2(n+1) \\ 
\therefore \quad c=1990, d=3978 .
\end{aligned}
\end{array}
\]