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Math Help - Unboundedness and convergence

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    Unboundedness and convergence

    what would be an example of a convergent subsequence of an unbounded sequence?
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    a_n  = \left\{ {\begin{array}{lr}<br />
   {n,} & {n \text{ is even}}  \\<br />
   {\frac{1}<br />
{n},} & \text{otherwise}  \\<br /> <br />
 \end{array} } \right.
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    ah yes! thank you
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    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.
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    How would you go on to prove that is unbounded? I can only figure out how to show it is convergent..
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by CrazyCat87 View Post
    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|
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    a_n  = \left\{ {\begin{array}{lr}<br />
   {n,} & {n \text{ is even}}  \\<br />
   {\frac{1}<br />
{n},} & \text{otherwise}  \\<br /> <br />
 \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?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by CrazyCat87 View Post
    a_n  = \left\{ {\begin{array}{lr}<br />
   {n,} & {n \text{ is even}}  \\<br />
   {\frac{1}<br />
{n},} & \text{otherwise}  \\<br /> <br />
 \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.
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    Quote Originally Posted by CrazyCat87 View Post
    a_n  = \left\{ {\begin{array}{lr}<br />
   {n,} & {n \text{ is even}}  \\<br />
   {\frac{1}<br />
{n},} & \text{otherwise}  \\<br /> <br />
 \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?
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  10. #10
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    Quote Originally Posted by CrazyCat87 View Post
    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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