16.2.25 ** Arrange natural numbers in sequence:
$$
123456789101112131415161718192021 \cdots \cdots
$$

If the $10^{n}$-th digit is in an $m$-digit number, then define $f(n)=m$. For example, the $10^{1}$-th digit 1 is in the two-digit number 10, and the $10^{2}$-th digit 5 is in the two-digit number 55, so $f(1)=2, f(2)=2$. Try to find $f(1991)$, and provide a proof.
Since there are $10^{r}-10^{r-1}$ numbers with $r$ digits. Let $g(r)$ be the total number of digits written up to the $r$-digit numbers in (*), then
$$
\begin{aligned}
g(r)= & \sum_{k=1}^{r} k\left(10^{k}-10^{k-1}\right)=-1+\sum_{k=1}^{r-1}[k-(k+1)] \cdot 10^{k} \\
& +r \cdot 10^{r}=-\sum_{k=0}^{r-1} 10^{k}+r \cdot 10^{r} \\
= & r \cdot 10^{r}-\frac{10^{r}-1}{9}, r \geqslant 1 .
\end{aligned}
$$

But $01987 \cdot 10^{1988}>10^{3} \cdot 10^{1988}=10^{1991}$, hence $f(1991)=1988$.