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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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    It's round donwn or truncate. And yes, @André Nicolas, I'm looking for a formal argument, thanks a lot.2012-04-01

3 Answers 3

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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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    @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.2012-04-01
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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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    @Théophile: Yes, I was careless i$n$ my comme$n$t, which has been deleted. I meant the *distance* to the nearest integer.2012-04-01
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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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    $M_{\epsilon}$ is just some sufficiently large natural number. I write $\epsilon$ in the subscript to indicate that this $M$ is a function of $\epsilon$.2012-04-01