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Is this claim true?

Given $\lim \limits_{n\to \infty}\ a_n=\frac{1}{2}$ Then $\lim \limits_{n\to \infty}\ (a_n - [a_n])=\frac{1}{2}$

I think it's true, but probably I just didn't find the right example to disprove it.

Thanks a lot.

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What do you mean by $[a_{n}]$? –  Isaac Solomon Apr 1 '12 at 18:14
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it's the round integer –  Anonymous Apr 1 '12 at 18:17
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Round up or down? –  Isaac Solomon Apr 1 '12 at 18:17
    
If you mean nearest integer, then the claim is correct, so you will not find an example to disprove it. Are you looking for a formal argument? –  André Nicolas Apr 1 '12 at 18:20
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It's round donwn or truncate. And yes, @André Nicolas, I'm looking for a formal argument, thanks a lot. –  Anonymous Apr 1 '12 at 18:21
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3 Answers

up vote 2 down vote accepted

Since

$$\lim_{n \to \infty} a_{n} = \frac{1}{2}$$

there exists $N$ such that for all $n \geq N$,

$$|a_{n} - \frac{1}{2}| < \frac{1}{4}$$

Then, for $n \geq N$, we have

$$0 \leq a_{n} \leq 1 \Longrightarrow [a_{n}] = 0$$

so that for $n \geq N$,

$$a_{n} - [a_{n}] = a_{n}$$

Now pick $\epsilon > 0$. Suppose that for all $n > M_{\epsilon}$

$$|a_{n} - \frac{1}{2}| < \epsilon$$

Replacing $M_{\epsilon}$ with $\mbox{max}(M_{\epsilon},N)$, we have

$$|(a_{n} - [a_{n}]) - \frac{1}{2} | = |a_{n} - \frac{1}{2}| < \epsilon$$

which proves that

$$\lim_{n \to \infty} (a_{n} - [a_{n}]) = \frac{1}{2}$$

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I'm reading the proof and I'm at the line: "Now pick $\epsilon > 0$. Suppose that for all $n > M_{\epsilon}$ - what is $M_{\epsilon}$? –  Anonymous Apr 1 '12 at 18:43
    
$M_{\epsilon}$ is just some sufficiently large natural number. I write $\epsilon$ in the subscript to indicate that this $M$ is a function of $\epsilon$. –  Isaac Solomon Apr 1 '12 at 19:29
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It depends on what you mean by the symbol $[x]$. If $[x]$ is the floor of $x$ (i.e., the greatest integer less than or equal to $x$), then the limit will be $\frac{1}{2}$. This follows from $\lim \limits_{n\to \infty}\ a_n=\frac{1}{2}$: there exists an $N$ such that for every $n \geq N$, $|a_n - \frac{1}{2}| < \frac{1}{2}$. For these $a_n$, it follows that $[a_n] = 0$, and so $$\lim \limits_{n\to \infty}\ a_n - [a_n] =\frac{1}{2} - 0 = \frac{1}{2}$$

On the other hand, if $[x]$ is the nearest integer to $x$, then the limit does not exist: consider the sequence $a = (0.4,0.6,0.49,0.51,0.499,0.501,\ldots)$. Clearly $a_n \rightarrow \frac{1}{2}$, but $[a_n]$ alternates between $0$ and $1$.

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Thanks, I'm talking about the first case where $[x]$ is the floor of $x$. Could you please explain why it follows that $[a_n] = 0$? –  Anonymous Apr 1 '12 at 18:32
    
Essentially, the $a_n$ are getting closer and closer to $\frac{1}{2}$; after a while (for $n \geq N$), all the $a_n$ will be between $0$ and $1$. The floor of a number in this interval is $0$. –  Théophile Apr 1 '12 at 18:39
    
@Théophile: Yes, I was careless in my comment, which has been deleted. I meant the distance to the nearest integer. –  André Nicolas Apr 1 '12 at 18:56
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We want to show that for any $\epsilon>0$, there is an $N$ such that if $n>N$, then $|(a_n-\lfloor a_n\rfloor)-\frac{1}{2}|<\epsilon$. Here by $\lfloor w\rfloor$ we mean the greatest integer which is $\le w$.

Let $\epsilon'=\min(\epsilon,1/4)$. By the fact that $\lim_{n\to\infty} a_n=\frac{1}{2}$, there is an $N$ such that if $n>N$ then $|a_n-\frac{1}{2}|<\epsilon'$. In particular, $|a_n-\frac{1}{2}|<\frac{1}{4}$, and therefore $\lfloor a_n\rfloor=0$. It follows that if $n>N$ then $$\left|(a_n-\lfloor a_n\rfloor)-\frac{1}{2}\right|=\left|a_n-\frac{1}{2}\right|<\epsilon.$$

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you wrote: $\left|(x_n-\lfloor x_n\rfloor)-\frac{1}{2}\right|=\left|x_n-\frac{1}{2}\right|<\epsilon$. But actually I know that $\left|(x_n-\lfloor x_n\rfloor)-\frac{1}{2}\right|=\left|x_n-\frac{1}{2}\right|<\epsilon'$, no? why can I make the switch from $\epsilon'$ to $\epsilon$? at least not for that n, maybe we need to add $n_1$? –  Anonymous Apr 1 '12 at 19:04
    
The number we are "challenged with" is $\epsilon$. If $\epsilon$ is ridiculous, like $5$, then we can't say that $\lfloor x_n\rfloor=0$. Specifying that $\epsilon'=\min(1/4,\epsilon)$ makes sure $\lfloor x_n\rfloor=0$. It also makes sure (since $\epsilon' \le \epsilon$) that from $|x_n-\frac{1}{2}|<\epsilon'$, we can conclude that $|x_n-\frac{1}{2}|<\epsilon$. More informally, we could forget about $\epsilon'$, and say that the argument will only work if $\epsilon$ is reasonably small, but that's good enough. –  André Nicolas Apr 1 '12 at 19:16
    
Thanks, I used the max between the n you gave me and the $n_1$ I have from the definition of limit, therefore safe to any $\epsilon$. –  Anonymous Apr 1 '12 at 19:23
    
@Anonymous: It is just a technical trick. It can be replaced by taking $N$ simultaneously large enough to make $|a_n-1/2|<1/4$ (anything under $1/2$ is OK) and large enough to make $|a_n-1/2|<\epsilon$. –  André Nicolas Apr 1 '12 at 19:27
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