- Thread starter
- #1

- Jan 17, 2013

- 1,667

I mean can we use a contradiction or show that two sub-sequences have a different limit ?

- Thread starter ZaidAlyafey
- Start date

- Thread starter
- #1

- Jan 17, 2013

- 1,667

I mean can we use a contradiction or show that two sub-sequences have a different limit ?

- Feb 13, 2012

- 1,704

A sequence $a_{n}$ that admits limit l, i.e. for an $\varepsilon > 0$ there exists an integer N so that for any n> N is $|a_{ n}-l| < \varepsilon$, converges. If l doesn't exist the sequence $a_{n}$ diverges... very simple concept?... yes, but... there are some little controversial about it...

I mean can we use a contradiction or show that two sub-sequences have a different limit ?

Kind regards

$\chi$ $\sigma$

We will show that $\cos(n), n = 0, 1, 2, \dots$ is dense in [-1,1], so the sequence $\cos(n)$ cannot converge.

Lemma: If $\theta$ is irrational, then the set $\{n \theta \pmod{1}: n \in \mathbb{Z}\}$ is dense in [0,1].

Proof: Let $\epsilon > 0$, and choose $k$ so that $0 < 1/k < \epsilon$. Consider the $k+1$ distinct values $\{i \theta \pmod{1}: i = 0, 1, \dots ,k\}$ and the $k$ intervals $[0, 1/k), [1/k, 2/k), \dots [(k-1)/k, 1)$. Since there are $k+1$ values, each of which is in one of the $k$ intervals, by the pigeonhole principle one of the intervals must contain at least two of the values, say $i_1 \theta \pmod{1}$ and $i_2 \theta \pmod{1}$ with $i_2 < i_1$. Then $0 < (i_2 - i_1) \theta \pmod{1} < 1/k$, so the values $m (i_2 - i_1) \theta \pmod{1}, m = 0, 1, 2, \dots ,M$ for some $M$, split [0,1] into intervals of width less than $1/k$. Every point in [0,1] must lie in one of these intervals and must therefore be within $1/k$ of $m (i_2 - i_1) \theta \pmod{1}$ for some $m$. This proves the lemma.

Now let $x \in [0,1]$, and apply the lemma to the irrational number $1/(2 \pi)$. By the lemma, there is a sequence of integers $m_i$ and $n_i$, $i = 1, 2, 3, \dots$ such that

$$\frac{m_i}{2 \pi} + n_i \to x \text{ as } i \to \infty$$

Then

$$m_i + 2 \pi n_i \to 2 \pi x$$

so by continuity of $\cos(x)$,

$$\cos(m_i) = \cos(m_i + 2 \pi n_i) \to \cos(2 \pi x)$$

Since $x$ is arbitrary in [0,1], this shows $\cos(n)$ is dense in [-1,1].

[edit] I added a factor of $2 \pi$ in one of the equations above.[/edit]

Lemma: If $\theta$ is irrational, then the set $\{n \theta \pmod{1}: n \in \mathbb{Z}\}$ is dense in [0,1].

Proof: Let $\epsilon > 0$, and choose $k$ so that $0 < 1/k < \epsilon$. Consider the $k+1$ distinct values $\{i \theta \pmod{1}: i = 0, 1, \dots ,k\}$ and the $k$ intervals $[0, 1/k), [1/k, 2/k), \dots [(k-1)/k, 1)$. Since there are $k+1$ values, each of which is in one of the $k$ intervals, by the pigeonhole principle one of the intervals must contain at least two of the values, say $i_1 \theta \pmod{1}$ and $i_2 \theta \pmod{1}$ with $i_2 < i_1$. Then $0 < (i_2 - i_1) \theta \pmod{1} < 1/k$, so the values $m (i_2 - i_1) \theta \pmod{1}, m = 0, 1, 2, \dots ,M$ for some $M$, split [0,1] into intervals of width less than $1/k$. Every point in [0,1] must lie in one of these intervals and must therefore be within $1/k$ of $m (i_2 - i_1) \theta \pmod{1}$ for some $m$. This proves the lemma.

Now let $x \in [0,1]$, and apply the lemma to the irrational number $1/(2 \pi)$. By the lemma, there is a sequence of integers $m_i$ and $n_i$, $i = 1, 2, 3, \dots$ such that

$$\frac{m_i}{2 \pi} + n_i \to x \text{ as } i \to \infty$$

Then

$$m_i + 2 \pi n_i \to 2 \pi x$$

so by continuity of $\cos(x)$,

$$\cos(m_i) = \cos(m_i + 2 \pi n_i) \to \cos(2 \pi x)$$

Since $x$ is arbitrary in [0,1], this shows $\cos(n)$ is dense in [-1,1].

[edit] I added a factor of $2 \pi$ in one of the equations above.[/edit]

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