3. Let $\left\{a_{n}\right\}$ be a sequence of positive numbers, and let the sum of the first $n$ terms be $b_{n}$. The product of the first $n$ terms of the sequence $\left\{b_{n}\right\}$ is $c_{n}$, and $b_{n}+c_{n}=1$. Then the number in the sequence $\left\{\frac{1}{a_{n}}\right\}$ that is closest to 2009 is $\qquad$.
3. 1980 Detailed Explanation: $b_{n}=\frac{c_{n}}{c_{n-1}}$, and $b_{n}+c_{n}=1$, thus $\frac{c_{n}}{c_{n-1}}+c_{n}=1$, which means $\frac{1}{c_{n}}-\frac{1}{c_{n-1}}=1$. From $c_{1}=b_{1}, c_{1}+b_{1}=1$ we get $c_{1}=b_{1}=\frac{1}{2}$. Therefore, $c_{n}=\frac{1}{n+1}, b_{n}=\frac{n}{n+1}$, and $\frac{1}{a_{n}}=n(n+1)$. The number in the sequence $\left\{\frac{1}{a_{n}}\right\}$ closest to 2009 is $44 \times 45=$ 1980.