Proof of a limit involving [x]

**redsoxfan325** · April 4th 2009, 03:00 PM

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.

**Chris L T521** · April 4th 2009, 03:51 PM

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, $\left[1\right]=\left\lfloor 1\right\rfloor=\boxed{1}$

**NonCommAlg** · April 4th 2009, 03:52 PM

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.

**redsoxfan325** · April 4th 2009, 05:36 PM

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 $[x_n]=0 ~\forall~ n$ .

**Moo** · April 5th 2009, 01:48 AM

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.

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