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Proves for the sequence $a_n$ with $a_n = \cos\left(\frac{n\pi}{4}\right)$

a)Show that it exist an $m \in \mathbb{N}$ that $a_n=a_{n+m} \forall n \in \mathbb{N}$ and determine $M=\{a_n ; n \in \mathbb{N}\}$
b) Find $\forall a \in M$ a subsequence $\{b_j\}$ of $\{a_n\}$ with $b_j = a \forall j \in \mathbb{N}$.Justify your answer.

c)Does a convergent subsequence $\{b_j\}$ of $\{a_n\}$ with $\{b_j; j \in \mathbb{N}\}=M$ exist?Justify your answer.

a)If $m=0: a_n =a_{n+m}$
If $n=1,m=6: a_n =a_{n+m}$
Do I only have to calculate $ \cos\left(\frac{\pi}{4}\right)= \frac{\sqrt{2}}{2}$ to define M?

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ok, thx for the advise. – phil Dec 20 '12 at 21:35
up vote 1 down vote accepted

Hint: Try evaluating/calculating $a_n$ for $n=1,2,3,...$ until you spot a pattern. The answers to all of the questions depend on this pattern.

I would guess that either the definition of $\mathbb{N}$ being used, or the intention of the question, means that $m=0$ is not considered a valid answer to part (a). The pattern that you should be able to spot will show you the answer. You should find a value of $m$ that works for all $n$: $m=6$ works for $n=1$, but not for $n=2$.

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Ok so I found for m=8 is the periodicity.and $M=[-1,1]$ is the interval? – phil Dec 20 '12 at 21:32
That's right: $m=8$. I think $M$ should be a finite set rather than an interval. – user12477 Dec 20 '12 at 21:39
I think I haven't really understood what M is...M is every number for $a_n$ with $n \in N$ is that right? – phil Dec 20 '12 at 21:46
Think of $M$ as giving you all possible values that occur in the list of numbers $a_1, a_2, a_3,...$ – user12477 Dec 20 '12 at 22:01
Well, then M = {pi/4;pi/2;3/4*pi;pi;5pi/4;3pi/2;7pi/2;2pi} because that are all the values for a_n with $n \in \mathbb{N_0}$, right? – phil Dec 21 '12 at 3:01

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