$n^3=n.n^2>n(n^2-1)=(n-1)n(n+1)\\ =>\dfrac{1}{n^3} < \dfrac{1}{(n-1)n(n+1)} \\ \dfrac{1}{(n-1)n(n+1)} =\dfrac{1}{2}.\dfrac{n+1-(n-1)}{(n-1)n(n+1)}=\dfrac{1}{2}\left(\dfrac{1}{(n-1)n}-\dfrac{1}{n(n+1)}\right)\\ =>\dfrac{1}{3^3}+\dfrac{1}{5^3}+...+\dfrac{1}{2009^3}\\ <\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{2008.2009.2010}\\ =\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{2008.2009}-\dfrac{1}{2009.2010}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{2009.2010}\right)\\ =\dfrac{1}{12}-\dfrac{1}{2.2009.2010}\\ <\dfrac{1}{12}$
