Example 1. (1989, American Mathematical Competition) For every positive integer $n$, let $S_{4}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$, $T_{n}=S_{1}+S_{2}+\cdots+S_{n}, U_{a}=\frac{1}{2} T_{1}+\frac{1}{3} T_{2}+\cdots+\frac{1}{n+1} T_{n}$. Try to find integers $0<a, b, c, d<1000000$, such that $T_{1088}=a \cdot S_{1988}-b, U_{1988}=c \cdot S_{1980}-d$, and prove your result.
Let $S_{0}=0$, then by Abel's identity, we have
$$
\begin{aligned}
T_{\mathrm{n}} & =\sum_{\mathrm{i}=0}^{\mathrm{n}} S_{1}=\sum_{k=0}^{\mathrm{n}-1}\left(\sum_{\mathrm{i}=0}^{\mathrm{k}} 1\right) \\
& \cdot\left(S_{\mathrm{k}}-S_{k+1}\right)+\left(\sum_{i=0}^{\mathrm{n}} 1\right) S_{\mathrm{n}} \\
& =\sum_{\mathbf{k}=0}^{\mathrm{n}-1}(k+1) \cdot\left(-\frac{1}{k+1}\right)+(n+1) S_{\mathrm{n}} \\
& =(n+1) S_{\mathrm{n}}-n \\
& =(n+1) S_{\mathrm{n}+1}-(n+1),
\end{aligned}
$$

i.e., $T_{\mathrm{n}}=(n+1) S_{\mathrm{n}+1}-(n+1)$.
Therefore, $a=b=1989$.
Also, for $T_{1}=(i+1)\left(S_{1+1}-1\right)(i=1$, $2, \cdots), 7:$ we have,
$$
\begin{aligned}
U_{n} & =\sum_{1=1}^{n} \frac{1}{i+1} T_{1}=\sum_{1=1}^{n}\left(S_{1+1}-1\right) \\
& =\sum_{1=1}^{n+1}\left(S_{1}-1\right)=\eta_{n+1}-(n+2) .
\end{aligned}
$$

From (2), we know
$$
U_{u}=(n+2) S_{n+1}-2(n+1),
$$

thus $c=1.990, d=2 \times 1989=3978$.