floor function sum

2009?

Since all these terms are between 1 and 2, each term can be taken as 1+{x} where {x} is the frac part.
So √2 = 1+ {√2 } etc.
So A = 2009 + {√2 } + {³√3/2 } + .....
So [A] = 2009 + [ {√2 } + {³√3/2 } + ..... + ]
I dunno wat to do next..

I just thought that after this (after sri's step) the sum in the box function on the right always <1.

Well even I assumed that sum to be < 1. Each term in the sum is lesser than the preceding term but I still dont understand why the sum has to be < 1.