# Thread: Proof of a limit involving [x]

1. ## Proof of a limit involving [x]

Prove or disprove: $\lim_{n\to\infty} \left[\sum_{k=1}^n \frac{1}{2^k}\right] = 1$, where $[x]$ is the greatest integer function.

2. Originally Posted by redsoxfan325
Prove or disprove: $\lim_{n\to\infty} \left[\sum_{k=1}^n \frac{1}{2^k}\right] = 1$, where $[x]$ is the greatest integer function.
This the same as $\left[\sum_{k=1}^{\infty}\left(\frac{1}{2}\right)^k\righ t]=\left[\frac{1}{1-\frac{1}{2}}-1\right]=\left[1\right]$

The greatest integer function of x is the same thing as the floor of x. Thus, $\left[1\right]=\left\lfloor 1\right\rfloor=\boxed{1}$

3. Originally Posted by redsoxfan325
Prove or disprove: $\lim_{n\to\infty} \left[\sum_{k=1}^n \frac{1}{2^k}\right] = 1$, where $[x]$ is the greatest integer function.
for any positive integer $n: \ \left[\sum_{k=1}^n \frac{1}{2^k}\right] = 0,$ because $0 < \sum_{k=1}^n \frac{1}{2^k} < 1.$ so the limit is 0 not 1.

4. Originally Posted by NonCommAlg
for any positive integer $n: \ \left[\sum_{k=1}^n \frac{1}{2^k}\right] = 0,$ because $0 < \sum_{k=1}^n \frac{1}{2^k} < 1.$ so the limit is 0 not 1.
Yeah, you're definitely right, now that I think about it. Let $\{x_n\}$ be a sequence where $x_n = \left[\sum_{k=1}^n \frac{1}{2^k}\right]$. In order for this sequence (and therefore the limit) to converge to 1, $\forall~ \epsilon>0, \exists~ N$ such that $\forall~ n>N, |1-x_n|<\epsilon$, and this is clearly false, as $\sum_{k=1}^n \frac{1}{2^k} < 1$, so $[x_n]=0 ~\forall~ n$.

5. Originally Posted by Chris L T521
This the same as $\left[\sum_{k=1}^{\infty}\left(\frac{1}{2}\right)^k\righ t]=\left[\frac{1}{1-\frac{1}{2}}-1\right]=\left[1\right]$

The greatest integer function of x is the same thing as the floor of x. Thus, $\left[1\right]=\left\lfloor 1\right\rfloor=\boxed{1}$
You can't put the limit inside the integer function.

$\sum_{k=1}^n \frac{1}{2^k}=\frac 12 \cdot \frac{1-\frac{1}{2^n}}{1-\frac 12}=1-\frac{1}{2^n}$, 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.