Unboundedness and convergence

**CrazyCat87** · January 15th 2010, 02:58 PM

what would be an example of a convergent subsequence of an unbounded sequence?

**Plato** · January 15th 2010, 03:07 PM

$a_n = \left\{ {\begin{array}{lr} {n,} & {n \text{ is even}} \\ {\frac{1} {n},} & \text{otherwise} \\ \end{array} } \right.$

**CrazyCat87** · January 15th 2010, 03:08 PM

ah yes! thank you

**HallsofIvy** · January 16th 2010, 12:44 AM

Here, is my first thought:
1, 2, 1, 3, 1, 4, 1, 5, 1, 6, ...

i.e. $a_n= 1$ if n is odd, $a_n= n/2$ if n is even. Basically, just choose any convergent and divergent series you want and "interleave" them.

**CrazyCat87** · January 16th 2010, 01:34 PM

How would you go on to prove that is unbounded? I can only figure out how to show it is convergent..

**Drexel28** · January 16th 2010, 08:13 PM

Originally Posted by CrazyCat87

what would be an example of a convergent subsequence of an unbounded sequence?

$a_n=n\left|\sin\left(\tfrac{\pi n}{2}\right)\right|$

**CrazyCat87** · January 20th 2010, 07:22 PM

$a_n = \left\{ {\begin{array}{lr} {n,} & {n \text{ is even}} \\ {\frac{1} {n},} & \text{otherwise} \\ \end{array} } \right.$

How do you prove this is unbounded? Is it enough if I prove that it is divergent and therefore unbounded? is there an easier way?

**Drexel28** · January 20th 2010, 07:29 PM

Originally Posted by CrazyCat87

$a_n = \left\{ {\begin{array}{lr} {n,} & {n \text{ is even}} \\ {\frac{1} {n},} & \text{otherwise} \\ \end{array} } \right.$

How do you prove this is unbounded? Is it enough if I prove that it is divergent and therefore unbounded? is there an easier way?

A sequence is really just a function defined on the natural numbers. And so you have a sequence $f:\mathbb{N}\mapsto\mathbb{Q}$ given by $f(n)=\begin{cases} n & \mbox{if} \quad n\text{ is even} \\ \frac{1}{n} & \mbox{if}\quad n\text{ is odd}\end{cases}$ . A sequence is bounded then if $\text{diam}\text{ }f\left(\mathbb{N}\right)$ is bounded. This is clearly not the case here.

**HallsofIvy** · January 21st 2010, 06:40 AM

Originally Posted by CrazyCat87

$a_n = \left\{ {\begin{array}{lr} {n,} & {n \text{ is even}} \\ {\frac{1} {n},} & \text{otherwise} \\ \end{array} } \right.$

How do you prove this is unbounded? Is it enough if I prove that it is divergent and therefore unbounded? is there an easier way?

Surely, you know the meaning of "undbounded", don't you? Look at the sequence for n even. Does that have an upper bound?

**drumist** · January 21st 2010, 07:00 AM

Originally Posted by CrazyCat87

How do you prove this is unbounded? Is it enough if I prove that it is divergent and therefore unbounded? is there an easier way?

Show that given any $M \in \Re$ , there exists an element of the sequence that is greater than $M$ .

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