1. Let $a_{n}=\frac{2^{n}}{2^{2 n+1}-2^{n+1}-2^{n}+1}(n$ be a positive integer). Then the value of $a_{1}+a_{2}+\cdots+a_{2011}$ is $\qquad$ 1 (fill in “greater than”, “equal to”, or “less than”).
$$
\begin{array}{l}
\text { Given } a_{n}=\frac{2^{n}}{\left(2^{n}-1\right)\left(2^{n+1}-1\right)} \\
=\frac{1}{2^{n}-1}-\frac{1}{2^{n+1}-1},
\end{array}
$$
$$
\begin{array}{l}
\text { we get } a_{1}+a_{2}+\cdots+a_{2011} \\
=\left(1-\frac{1}{2^{2}-1}\right)+\left(\frac{1}{2^{2}-1}-\frac{1}{2^{3}-1}\right)+\cdots+ \\
\left(\frac{1}{2^{2011}-1}-\frac{1}{2^{2012}-1}\right) \\
=1-\frac{1}{2^{2012}-1}<1 .
\end{array}
$$