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How to determine if a sequence is Inc, Dec, or not monotonic? Aso if its bounded?

  1. Nov 28, 2005 #1
    Need help clairfying some stuff.
    How do you determine if a Sequence is not monotonic? Also if its just inc. or dec. its monotonic?
    For example.
    Seq=An= 1/(2n+3)
    First 4 terms are {1/5,1/7,1/9,1/11,...}
    So its decreasing...and I guess monotonic?
    And how would you determine if that sequences is bounded?
    It looks like a upper bound of 1/5 but a lower bound of 0?
    Help?
     
  2. jcsd
  3. Nov 28, 2005 #2
    yes, if it's decreasing then it's monotonic. i think if you look in your book the definition of decreasing is something like "a sequence is decreasing if a_n > a_(n+1) or a_n = a_(n+1)" if you plug n & (n+1) into your sequence you'll see that it's decreasing. to show that a sequence isn't monotonic you'll have to show that it isn't decreasing nor increasing. so it would have to decrease & then increase, or increase then decrease


    a sequence is bounded if it fits the definition of a bounded sequence. can't remember what that is, you'll have to look it up.
     
  4. Nov 29, 2005 #3

    matt grime

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    in the case of that sequence, to show bounded ness, you just DO IT.

    all the terms are positve, hence bounded below by 0 (and -1, and -pi), and all of the terms are less than 1 hence bounded above by 1, and many other numbers. you don't have to find the 'best' bound.

    you'd have no problem showing that the sequence 2n+3 was strictly increasing and bounded below by 5, would you? so it's as easy to show the reciprocal is strictly decreasing and bounded above by 1/5
     
  5. Nov 29, 2005 #4

    HallsofIvy

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    If you want a really formal proof that a sequence is, say, decreasing and bounded below you can try this:

    If a sequence is decreasing- that is, [itex]a_{n+1}\leq a_n[/itex], then [itex]a_{n+1}- a_n\leq 0[/itex]. In this particular case,
    [tex]a_{n+1}- a_n= \frac{1}{2(n+1)+3}- \frac{1}{2n+3}= \frac{1}{2n+5}- \frac{1}{2n+3}= \frac{2n+3}{(2n+5)(2n+3)}- \frac{2n+5}{(2n+5)(2n+3)}= \frac{-2}{(2n+5)(2n+3)}[/tex]
    which is clearly negative.

    Similarly, if [itex]a_{n+1}\leq a_n[/itex], then [itex]\frac{a_{n+1}}{a_n}\lequ 1[/itex].
    In this case [itex]\frac{a_{n+1}}{a_n}= \frac{\frac{1}{2n+5}}{\frac{1}{2n+3}}= \frac{1}{2n+5}\frac{2n+3}{1}= \frac{2n+3}{2n+5}[/tex]
    which is clearly less than 1.

    Every number in that sequence is positive so it has 0 as a lower bound. By the "monotone convergence property" (which what this is all about), that sequence must converge to some number larger than or equal to 0. In fact, here it is obvious that the limit of the sequence is 0.
     
  6. Feb 6, 2010 #5
    Hi dose any have any examples of non monotonic sequences .....

    please help
     
  7. Feb 6, 2010 #6
    a_n = sin(n)

    or

    a_n=(-1)^n
     
  8. Feb 7, 2010 #7
    i have to find a bounded sequence that consists of negitive no.s thats non monotonic ......any ideas am i am tearing my hair out

    thanks for the reply xx
     
  9. Feb 10, 2011 #8
    -2 + (-1)^n always negative and keeps wavering between -3, and -1.
     
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