Sequence question

**chella182** · May 6th 2009, 01:24 PM

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 $1\leq n_{1}<n_{2}<n_{3}<...$ , 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.

**Plato** · May 6th 2009, 01:39 PM

Originally Posted by chella182

We define a sequence by $x_{n}=\cos{(n\pi/3)}$ . Find an increasing list of numbers $1\leq n_{1}<n_{2}<n_{3}<...$ , 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 $\left( {\forall j} \right)\left[ {\cos \left( {\frac{{n_j \pi }}{3}} \right) = ?} \right]$

**chella182** · May 6th 2009, 01:42 PM

Wow, no idea.

**Plato** · May 6th 2009, 01:49 PM

Originally Posted by chella182

Wow, no idea.

What does that mean? Can you do the basic caculations?

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

**chella182** · May 6th 2009, 02:00 PM

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.

**Plato** · May 6th 2009, 02:38 PM

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.

**chella182** · May 6th 2009, 02:47 PM

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 $1\leq n_{1}<n_{2}<n_{3}<...$ , 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.

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