jmc

We are asked to find $$\frac{2}{1\cdot 3}+\frac{2}{2\cdot 4}+\frac{2}{3\cdot 5}+\frac{2}{4\cdot 6}+\cdots +\frac{2}{2009\cdot 2011}.$$ Observe that $\frac{2}{n(n+2)}$ may be written as $\frac{1}{n}-\frac{1}{n+2}$. Applying this identity, our sum becomes $$\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\frac{1}{4}-\frac{1}{6}+\cdots +\frac{1}{2009}-\frac{1}{2011}.$$ Every negative term cancels with the term three places to the right. The only terms which remain are $$1+\frac{1}{2}-\frac{1}{2010}-\frac{1}{2011}.$$ To the nearest thousandth, the sum is $\boxed{1.499}$.