Prove or disprove: $\displaystyle \lim_{n\to\infty} \left[\sum_{k=1}^n \frac{1}{2^k}\right] = 1$, where $\displaystyle [x]$ is the greatest integer function.
This the same as $\displaystyle \left[\sum_{k=1}^{\infty}\left(\frac{1}{2}\right)^k\righ t]=\left[\frac{1}{1-\frac{1}{2}}-1\right]=\left[1\right]$
The greatest integer function of x is the same thing as the floor of x. Thus, $\displaystyle \left[1\right]=\left\lfloor 1\right\rfloor=\boxed{1}$
Yeah, you're definitely right, now that I think about it. Let $\displaystyle \{x_n\}$ be a sequence where $\displaystyle x_n = \left[\sum_{k=1}^n \frac{1}{2^k}\right]$. In order for this sequence (and therefore the limit) to converge to 1, $\displaystyle \forall~ \epsilon>0, \exists~ N$ such that $\displaystyle \forall~ n>N, |1-x_n|<\epsilon$, and this is clearly false, as $\displaystyle \sum_{k=1}^n \frac{1}{2^k} < 1$, so $\displaystyle [x_n]=0 ~\forall~ n$.
You can't put the limit inside the integer function.
$\displaystyle \sum_{k=1}^n \frac{1}{2^k}=\frac 12 \cdot \frac{1-\frac{1}{2^n}}{1-\frac 12}=1-\frac{1}{2^n}$, which is obviously < 1.
Putting the limit inside the integer function omits the fact that 1 is a limit, not the value of the sum.