28. Consider a sequence of real numbers $\left\{a_{n}\right\}$ defined by $a_{1}=1$ and $a_{n+1}=\frac{a_{n}}{1+n a_{n}}$ for $n \geq 1$.
Find the value of $\frac{1}{a_{2005}}-2000000$.
28. Ans: 9011
For $n \geq 1, a_{n+1}=\frac{a_{n}}{1+n a_{n}} \Longrightarrow \frac{1}{a_{n+1}}=\frac{1+n a_{n}}{a_{n}}=\frac{1}{a_{n}}+n$. Therefore,
$$
\begin{aligned}
\frac{1}{a_{2000}}=\frac{1}{a_{2004}}+2004 & =\frac{1}{a_{2003}}+2003+2004 \\
& =\cdots \\
& =\frac{1}{a_{1}}+(1+2+3+\cdots+2004)=1+\frac{2004 \times 2005}{2}=2009011
\end{aligned}
$$

Hence, $\frac{1}{a_{2000}}-2000000=9011$.