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I need help with monotonic sequence

  1. Apr 22, 2013 #1
    1. Determine whether the sequence with the given nth term is monotonic & bounded.

    a_n = (n) / (2^(n+2))





    2.
    b_n < b_n+1




    3.

    (n) / (2^(n+2)) < (n+1) / (2^(n+3))

    I multiply both side by (2^(n+2)) and (2^(n+3))



    (n)(2^(n+3)) < (n+1)(2^(n+2))



    Then i distribute and got:

    (n)(2^(n+3)) < (n)(2^(n+2)) + (2^(n+2))



    this is where im stuck. where do i go from here?
     
    Last edited by a moderator: Apr 22, 2013
  2. jcsd
  3. Apr 22, 2013 #2

    Mark44

    Staff: Mentor

    You can't start with the inequality above, since that's what you want to end with if the sequence is monotonically decreasing. You can use the fact that 2n + 3 = 2 * 2n + 2.
     
  4. Apr 22, 2013 #3

    so just by knowing that 2n + 3 < 2 * 2n + 2 is good enough?
     
  5. Apr 22, 2013 #4

    Mark44

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    No, and it didn't say that 2n + 3 < 2 * 2n + 2.

    2n + 3 = 2 * 2n + 2
     
  6. Apr 22, 2013 #5
    okay. so whats next or is that it?
     
  7. Apr 22, 2013 #6

    LCKurtz

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    Below you are starting out assuming ##a_n < a_{n+1}## which, as Mark pointed out, you don't know. Maybe you could try ##a_n\, ?\, a_{n+1}## and leave the inequality unknown to explore and see where it leads.
    At that step why don't you try dividing out all the 2's you can. Maybe you can get to where you can decide if either ? = < or ? = > makes a true statement. Then what you need to do is start with that true statement and see if you can work backwards to what you are trying to prove.
     
  8. Apr 22, 2013 #7


    okay . so if i divide out the 2's then i get

    ((2^n+3) / (2^n+2)) ? (n+1) / (n)

    2 ? ((n+1) / n)

    at this point, im i doing it right
     
    Last edited by a moderator: Apr 22, 2013
  9. Apr 22, 2013 #8

    LCKurtz

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    You tell me. Have you simplified it enough that you know whether to use < or > ? What if you multiply both sides by n and keep simplifying?
     
    Last edited by a moderator: Apr 22, 2013
  10. Apr 22, 2013 #9
    i would get n ? 1
     
  11. Apr 22, 2013 #10

    LCKurtz

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    What about my other question?
     
  12. Apr 22, 2013 #11
    yes i have simplify enough.

    but how i do i know if its < or >?
     
  13. Apr 22, 2013 #12

    LCKurtz

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    OK. Now do you see how to start with "if ##n>1## then" and end with the statement that your sequence is monotone and whether it is increasing or decreasing?
     
  14. Apr 22, 2013 #13

    LCKurtz

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    Which one is true in your problem? n > 1 or n < 1? You need to start with a true statement and work backwards now.
     
  15. Apr 22, 2013 #14
    would i just plug in values for n which is n>1 and compare the values to see it its monotonic?
     
  16. Apr 22, 2013 #15
    0.125 > 0.093 > 0.0625 > 0.039 ...

    when i plug in values n > 1 into the function
     
  17. Apr 22, 2013 #16

    LCKurtz

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    I'm out of time and you obviously aren't getting the point of this discussion. Maybe Mark44 will take it from here. I have to go now.
     
  18. Apr 22, 2013 #17

    Mark44

    Staff: Mentor

    physics=world,
    Now that you have played with this for a while, what's your sense here?
    Is an + 1 > an?
    Or is an + 1 < an?
     
  19. Apr 22, 2013 #18
    an + 1 < an
     
  20. Apr 22, 2013 #19

    Mark44

    Staff: Mentor

    OK, so now let's see if we can prove it.

    I'm going to take a different tack than LCKurtz did, since you didn't seem to be following what he was doing.

    You would like to show that an + 1 < an

    Here's the start of the argument.
    $$ a_{n + 1} = \frac{n + 1}{2^{n+3}} =(1/2)\frac{n + 1}{2^{n + 2}} = (1/2)[\frac{n}{2^{n+2}} + \frac{1}{2^{n+2}}] $$

    We would eventually like to come out of this with "< an". What can you do with the right-most part of the equation above?
     
  21. Apr 22, 2013 #20
    can we simplify?
     
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