$\underset{n→∞}{\lim}\left[\sum\limits_{r=1}^{n}\frac{1}{2^r}\right]$, where [.] denotes the greatest integer function, is

equal to zero

$\sum\limits_{r=1}^{n}\frac{1}{2^r}=\frac{\frac{1}{2}\left(1-\left(\frac{1}{2}\right)^n\right)}{\left(1-\frac{1}{2}\right)}=1-\left(\frac{1}{2}\right)^n$, which tends to one as n → ∞ (but infact always remains
less than one). Thus $\underset{n→∞}{\lim}\left[\sum\limits_{r=1}^{n}\frac{1}{2^r}\right]=0$