Calculate \sum_{n=1}^{\infty}(\lfloor\sqrt[n]{2010}\rfloor-1) where \lfloor x\rfloor is the largest integer less than or equal to x.
Calculate \sum_{n=1}^{\infty}(\lfloor\sqrt[n]{2010}\rfloor-1) where \lfloor x\rfloor is the largest integer less than or equal to x.