1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Prove that sequence is monotonic

  1. Feb 15, 2013 #1
    1. The problem statement, all variables and given/known data
    Need to prove that the following sequence is monotonic ( decreasing ).

    [itex]\frac{1}{n^2}+\frac{(-1)^{n}}{3^n}[/itex]


    2. Relevant equations

    -

    3. The attempt at a solution

    I have idea how to prove that the sequence is decreasing that is:

    [itex]a_{n+1} - a_{n} ≤ 0[/itex]
    but in this case, I can't get the inequality proved.

    [itex]\frac{1}{(n+1)^2}+\frac{(-1)^{n+1}}{3^{n+1}} - \big( \frac{1}{n^2}+\frac{(-1)^{n}}{3^n} \big) ≤ 0[/itex]

    I end up in a mess, that looks like;

    [itex]\frac{3^{n+1}(-2n-1)-2n^{2}(n+1)^2(-1)^n}{n^2((n+1)^2)3^{n+1}}[/itex]

    Could get any hint which way I should start to reform the equation?

    Also I'm intrested in other thing. If I need to show that for example this sequence is bounded, how I can "see" what the limit "might " be.
    In every material I see, that they check the first few terms and make "good guess"?

    For example we have sequence [itex]a_0 = 0, a_{n+1} = (1/2)x_n +5[/itex]. We can see when calculating the first we values, that
    it seems to converge to [itex]x = 10[/itex]. But if the sequence is more complicated?
     
    Last edited: Feb 15, 2013
  2. jcsd
  3. Feb 15, 2013 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The 3th terms is 0.074074, the 4th term is 0.07484, greater. The sequence is not monotonous.

    ehild
     
  4. Feb 15, 2013 #3
    Ah, stupid me. Had calculated the first terms totally wrong.. Well, that could have saved some time and not bang my head against the wall! :D

    Moderators can lock this up, sorry for inconvience.
     
  5. Feb 15, 2013 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    To answer your second question: if ##a_{n+1} = (1/2)a_n + 5##, and IF ##a_n## has a limit ##L##, you can calculate ##L## from ##L = (1/2)L + 5##. In this case you can show that for any value of ##a_0## a limit does, indeed, exist. As for more complicated examples, sometimes the same type of 'trick' works (not always, but sometimes).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook