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Proving that a function is monotonic and bounded

  1. Dec 9, 2008 #1
  2. jcsd
  3. Dec 9, 2008 #2
    I think you may be making the monotone part a bit too complicated. Look at [tex]x_{n+1}/x_n[/tex]. What can you tell about this ratio?

    For the bounded part, why try to look at an upper bound for the numerator and a lower bound for the denominator?
     
  4. Dec 9, 2008 #3
    Your series is [tex]x_n=\prod_{i=1}^n \frac{i+9}{2i-1} [/tex]. I'll give you a hint, [tex]\frac{i+9}{2i-1}=\frac{1}{4} \frac{19}{2i-1}[/tex]. What happens as i increases (after i=5)?

    Good luck
     
  5. Dec 10, 2008 #4
    why i need to prove that An+1<An
    ??
     
  6. Dec 10, 2008 #5
    If you can show that A_(n+1)<A_n for n greater than some value, then it's decreasing for large enough n - which means its monotonic.
     
  7. Dec 10, 2008 #6

    Mark44

    Staff: Mentor

    Is there a typo in the line above? I don't see how this could be true.
     
  8. Dec 12, 2008 #7
    Yeah sorry that should be[tex] \frac{i+9}{2i-1}=\frac{1}{2}+\frac{1}{4} \frac{19}{2i-1}[/tex]
     
  9. Dec 12, 2008 #8

    Mark44

    Staff: Mentor

    I don't see that those two expressions are equal, either. On the right side you get 4i - 17 in the numerator, and 4(2i - 1) in the denominator.
     
  10. Dec 15, 2008 #9
    Yeah right again I miscalculated the factor [tex]\frac{i+9}{2i-1}=\frac{1}{2}+\frac{1}{2} \frac{19}{2i-1}[/tex]
     
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