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 $\{b_{n}\}$ be $c_{n}$, 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.

Given $b_{n}=\frac{c_{n}}{c_{n-1}}(n \geqslant 2), c_{1}=b_{1}=\frac{1}{2}$.
$$
\begin{array}{l}
\text { Then } \frac{c_{n}}{c_{n-1}}+c_{n}=1 \Rightarrow \frac{1}{c_{n}}-\frac{1}{c_{n-1}}=1 \\
\Rightarrow \frac{1}{c_{n}}=2+(n-1) \times 1=n+1 \\
\Rightarrow c_{n}=\frac{1}{n+1} \Rightarrow b_{n}=\frac{n}{n+1} \\
\Rightarrow a_{n}=b_{n}-b_{n-1}=\frac{n}{n+1}-\frac{n-1}{n}=\frac{1}{n(n+1)} \\
\Rightarrow \frac{1}{a_{n}}=n(n+1) .
\end{array}
$$

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.