${ }^{*} 7$. Find the smallest term in the sequence $\left\{a_{n}\right\}$. Where, $a_{\mathrm{n}}=n+\frac{1989}{n^{2}}$.
First, use a calculator to find the first few terms of the sequence $\left\{a_{\mathrm{n}}\right\}$:
$$
\begin{array}{l}
a_{1}=1990, a_{2}=499.25, a_{\mathrm{s}}=224, \\
\cdots, a_{15}=23.84, a_{18}=23.769531, \cdots, \\
a_{17}=23.882352, \cdots
\end{array}
$$

We conjecture: The smallest term of the sequence $\left\{a_{\mathrm{n}}\right\}$ is the number $a_{10}$. Let's prove this.
Consider the difference $b_{\mathrm{n}}=a_{\mathrm{a}+1}-a_{\mathrm{a}}$
$$
=1-1989 \cdot \frac{2 n+1}{n^{2}(n+1)^{2}} \text {. }
$$

The sequence $c_{0}=\frac{2 n+1}{n^{2}(n+1)^{2}}=\frac{2}{n(n+1)^{2}}$ $+\frac{1}{n^{2}(n+1)^{2}}$ is monotonically decreasing, because it is the sum of two monotonically decreasing sequences, so the general term $b_{\mathrm{a}}=1$ $-1989 c_{\mathrm{a}}$ is monotonically increasing. Since $b_{18}>0$; therefore, for all $n \geqslant 16$ we have $b_{n}>0$, i.e., for all $n \geqslant 16$, $a_{\mathrm{a}+1}>a_{\mathrm{a}}$. Therefore, the smallest term of the sequence $\left\{a_{0}\right\}$ is $a_{18}=23.769531 \cdots$.