Thread: Unboundedness and convergence

1. Unboundedness and convergence

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

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

3. ah yes! thank you

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

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

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

6. Originally Posted by CrazyCat87
what would be an example of a convergent subsequence of an unbounded sequence?
$\displaystyle a_n=n\left|\sin\left(\tfrac{\pi n}{2}\right)\right|$

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

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?

8. Originally Posted by CrazyCat87
$\displaystyle a_n = \left\{ {\begin{array}{lr} {n,} & {n \text{ is even}} \\ {\frac{1} {n},} & \text{otherwise} \\ \end{array} } \right.$

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 $\displaystyle f:\mathbb{N}\mapsto\mathbb{Q}$ given by $\displaystyle f(n)=\begin{cases} n & \mbox{if} \quad n\text{ is even} \\ \frac{1}{n} & \mbox{if}\quad n\text{ is odd}\end{cases}$. A sequence is bounded then if $\displaystyle \text{diam}\text{ }f\left(\mathbb{N}\right)$ is bounded. This is clearly not the case here.

9. Originally Posted by CrazyCat87
$\displaystyle a_n = \left\{ {\begin{array}{lr} {n,} & {n \text{ is even}} \\ {\frac{1} {n},} & \text{otherwise} \\ \end{array} } \right.$

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?

10. 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 $\displaystyle M \in \Re$, there exists an element of the sequence that is greater than $\displaystyle M$.