6. The graph of the function $f(x)=\sum_{k=1}^{2011} \frac{x+k-1}{x+k}$ is symmetric about the center at
6. $(-1006,2011)$.

Notice that,
$f(x)=\sum_{k=1}^{2011} \frac{x+k-1}{x+k}=2011-\sum_{k=1}^{2011} \frac{1}{x+k}$.
Let $g(x)=\sum_{k=1}^{2011} \frac{1}{x+k}$. Then
$$
g(x-1006)=\sum_{k=-1005}^{1005} \frac{1}{x+k}
$$

is an odd function.
Thus, the graph of $g(x)$ is symmetric about $(-1006,0)$.
Therefore, the graph of $f(x)$ is symmetric about $(-1006,2011)$.