jmc

Recall that ${\log }_2\frac{x}{y}={\log }_{2}x-{\log }_{2}y$. Applying this identity to each term in the sum, we find that the sum is equal to to $({\log }_{2}2-{\log }_{2}1)+({\log }_{2}3-{\log }_{2}2)+\cdots +({\log }_{2}2010-{\log }_{2}2009)$. Most of the intermediate terms cancel out; the expression eventually evaluates to $${\log }_{2}2010-{\log }_{2}1={\log }_{2}2010.$$Note that $2^{10}=1024$, but $2^{11}=2048$, so $10<{\log }_{2}2010<11$. It follows that the largest integer less than ${\log }_2\frac{2}{1}+{\log }_2\frac{3}{2}+\cdots +{\log }_2\frac{2009}{2008}+{\log }_2\frac{2010}{2009}$ is $\boxed{10}$.