1. ## Sequence question

First off, sorry if this is in the wrong section.
The question goes...

We define a sequence by $x_{n}=\cos{(n\pi/3)}$. Find an increasing list of numbers $1\leq n_{1}, so that the subsequence $y_{k}=x_{n_{k}}$ tends to 1. Prove that $\{x_{n}\}_{n=1}^{\infty}$ does not tend to a limit.

I have no clue. I hate stuff like this.

2. Originally Posted by chella182
We define a sequence by $x_{n}=\cos{(n\pi/3)}$. Find an increasing list of numbers $1\leq n_{1}, so that the subsequence $y_{k}=x_{n_{k}}$ tends to 1. Prove that $\{x_{n}\}_{n=1}^{\infty}$ does not tend to a limit.
Consider $n_k = 6\cdot k$. Then $\left( {\forall k} \right)\left[ {\cos \left( {\frac{{n_k \pi }}{3}} \right) = ?} \right]$

This time,
Consider $n_j = 9\cdot (2j+1)$. Then $\left( {\forall j} \right)\left[ {\cos \left( {\frac{{n_j \pi }}{3}} \right) = ?} \right]$

3. Wow, no idea.

4. Originally Posted by chella182
Wow, no idea.
What does that mean? Can you do the basic caculations?

$n_1=6,~n_2=12,~n_3=18,~...$
${\cos \left( {\frac{{n_1 \pi }}{3}} \right) = ?}$
${\cos \left( {\frac{{n_2 \pi }}{3}} \right) = ?}$
${\cos \left( {\frac{{n_3 \pi }}{3}} \right) = ?}$

5. Means I have no idea what you're talking about. Where are you getting the $n_{1}=6$ et cetera from? I just don't understand it at all.

6. Originally Posted by chella182
Means I have no idea what you're talking about. Where are you getting the $n_{1}=6$ et cetera from? I just don't understand it at all.
If that is true, then your difficulties extend far beyond any help you can realistically expect to get anywhere on a website. You need a good ole sit-down with a live instructor.

7. Okay, I have a similar example with a solution, but I don't understand it either.

We define a sequence by $x_{n}=\sin{(n\pi/2)}$.Find an increasing list of numbers $1\leq n_{1}, so that the subsequence $y_{k}=x_{n_{k}}$ tends to 0. Prove that $\{x_{n}\}^{\infty}_{n=1}$ does not tend to a limit.

The answer for the first bit says...

A subsequence where the value is zero is given by $n_{k}=2k$, so $2<4<6<8$...

Why $2k$ though?
The answer to the second bit it says...

To prove the sequenece does not tend to a limit, we have to find a subsequence where the limit is not zero. Take $n_{k}=4k+1$, then $\sin{((4k+1)\pi/2)}=\sin{(2k\pi+\pi/2)}=1$. Now we say, if the sequence $\{x_{n}_{n=1}^{\infty}$ tend to a limit $l$, then so do the subsequences, hence $0=l=1$, which is a contradiction, hence the assumption $\{x_{n}\}_{n=1}^{\infty}$ tends to a limit is false, and the sequence does not tend to a limit.

Why $n_{k}=4k+1$? And why does $\sin{(2k\pi+\pi/2)}=\sin{(\pi/2)}$? I just don't get it.