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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.2012-12-20

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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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    Actually it will correspond to the $\cos$ of the values you listed - think you forgot this bit. So you should see that $M=\{-1, -\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt 2},1\}$. To answer parts (b) and (c), try to pick out the subsequence $b_j$ that has (for example) $b_j=-\frac{1}{\sqrt{2}}$ for every $j$.2012-12-21