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Sequence integral help

  1. Jan 9, 2008 #1
    show that 1/n+1 +1/n+2 +....1/2n≤from 2n to n∫dt/t≤1/n+...1/2n-1



    consider the sequene (Vn) defined by:
    Vn= 1/n+1 +...+1/2n=from 2n to p=n+1∑1/p
    deduce from above that ln2-1/2n≤ Vn≤ln2




    i didnt find any thing
     
  2. jcsd
  3. Jan 9, 2008 #2

    HallsofIvy

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    Are you saying "Show that
    [tex]\frac{1}{n+1}+ \frac{1}{n+2}+ \cdot\cdot\cdot + \frac{1}{2n}\le \int_{2n}^n\int \frac{dt}{t}\le \frac{1}{n}+ \cdot\cdot\cdot + \frac{1}{2n-1}[/itex]

    Looks to me like the integral in the middle will be negative: [itex]\int_{2n}^n dt/t= ln(n)- ln(2n)= ln(n/2n)= ln(1/2)= -ln(2)= -0.693147[/itex] and that can't possibly statisfy the sums you give.


     
    Last edited: Jan 11, 2008
  4. Jan 9, 2008 #3
    [tex]\frac{1}{n+1}+ \frac{1}{n+2}+ \cdot\cdot\cdot + \frac{1}{2n}\le \int_{2n}^n\frac{dt}{t}\le \frac{1}{n}+ \cdot\cdot\cdot + \frac{1}{2n-1}[/tex]
    its like this
    not two integrals its one integral
     
  5. Jan 9, 2008 #4
    Chicky, by what you want to deduce it looks like you have your bounds of integration backwards
     
  6. Jan 11, 2008 #5
    [tex]\frac{1}{n+1}+ \frac{1}{n+2}+ \cdot\cdot\cdot + \frac{1}{2n}\le \int_{2n}^n\frac{dt}{t}\le \frac{1}{n}+ \cdot\cdot\cdot + \frac{1}{2n-1}[/tex]
    we must show this not deduce it
     
  7. Jan 11, 2008 #6
    Your inequalities aren't true, so you can't "show" this. What Kreizhn is trying to say that you may have your limits of integration in the wrong order (by switching them, the inequalities are then true). Anyway, here's what you should try: consider approximating the area underneath 1/t using n rectangles of width 1 (i.e. all rectangles have their upper right corners on 1/t). Then try approximating the integral by overestimating -- use n rectangles of width 1 but with their upper left corners on 1/t. If you graph 1/t and draw out the rectangles and compute their area, it will be fairly obvious how the inequalities work.
     
  8. Jan 11, 2008 #7
    am sorry the limit are in the wrong order but the inequality is true am sure
     
  9. Jan 11, 2008 #8
    Let me clarify. What I meant was: with the incorrect order (limits of integration) the inequalities would (obviously) not be true.

    Did you try my hint? This is just an exercise in Riemann sums.
     
  10. Jan 11, 2008 #9
    i just dont know it
    thanks any ways
     
  11. Jan 11, 2008 #10

    HallsofIvy

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    You don't know what a Riemann sum is? Or do you mean you just don't want to try what has been suggested?
     
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