(5) Let $\{a_{n}\}$ 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 $\{b_{n}\}$ 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 2011 is ( ).
(A) 1980
(B) 2010
(C) 2040
(D) 2070
(5) A Hint: Given $b_{n}=\frac{c_{n}}{c_{n-1}}(n \geqslant 2), c_{1}=b_{1}$, so $\frac{c_{n}}{c_{n-1}}+c_{n}=$ $1, c_{1}=b_{1}=\frac{1}{2}$. Therefore, $\frac{1}{c_{n}}-\frac{1}{c_{n-1}}=1$. So $\left\{\frac{1}{c_{n}}\right\}$ is an arithmetic sequence with the first term 2 and common difference 1.
So $\frac{1}{c_{n}}=2+(n-1) \cdot 1=n+1, c_{n}=\frac{1}{n+1}, b_{n}=\frac{n}{n+1}$. Therefore, $a_{n}=b_{n}-b_{n-1}=\frac{n}{n+1}-\frac{n-1}{n}=\frac{1}{n(n+1)}, \frac{1}{a_{n}}=n(n+1)$. Since $44 \times 45=1980,45 \times 46=2070$, the number in the sequence $\left\{\frac{1}{a_{n}}\right\}$ closest to 2011 is 1980.