11. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{n}=\sqrt{1+\frac{1}{n^{2}}+\frac{1}{(n+1)^{2}}}-1 \text {. }
$$

Then $a_{1}+a_{2}+\cdots+a_{2000}=$
11. $\frac{2009}{2010}$.

Notice that
$$
\begin{array}{l}
1+\frac{1}{k^{2}}+\frac{1}{(k+1)^{2}}=\frac{k^{4}+2 k^{3}+3 k^{2}+2 k+1}{k^{2}(k+1)^{2}} \\
=\left[\frac{k^{2}+k+1}{k(k+1)}\right]^{2} .
\end{array}
$$
$$
\begin{array}{l}
\text { Then } a_{k}=\sqrt{1+\frac{1}{k^{2}}+\frac{1}{(k+1)^{2}}}-1 \\
=\frac{k^{2}+k+1}{k(k+1)}-1=\frac{1}{k}-\frac{1}{k+1} .
\end{array}
$$

Thus, we have
$$
\begin{array}{l}
\sum_{k=1}^{2009} a_{k}=\sum_{k=1}^{2009}\left(\frac{1}{k}-\frac{1}{k+1}\right) \\
=1-\frac{1}{2010}=\frac{2009}{2010} .
\end{array}
$$