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Homework Help: Proof of a inequality

  1. Sep 29, 2010 #1
    1. The problem statement, all variables and given/known data

    Prove that

    1 + 1/2 + 1/4 + 1/7 + 1/11 + ............................ <= 2*pi


    2. Relevant equations
    none


    3. The attempt at a solution
    all i could figure out was the nth term of the sequence

    [tex]
    T(n) = \frac{2}{2 + n(n-1)}
    [/tex]

    any help appreciated.:biggrin:
     
  2. jcsd
  3. Sep 30, 2010 #2
    please give me a starting tip atleast !!! :approve:
     
  4. Sep 30, 2010 #3

    ehild

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  5. Sep 30, 2010 #4
  6. Sep 30, 2010 #5

    ehild

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    I do not have any other idea. What have you learnt about infinite series?

    ehild
     
  7. Sep 30, 2010 #6

    fzero

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  8. Sep 30, 2010 #7

    ehild

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    Try to approximate the circumference of the unit circle by inscribed polygons.

    ehild
     
  9. Sep 30, 2010 #8
    all i know is about diverging and converging GP (infinite) and sum of decreasing infinite GP.
     
  10. Sep 30, 2010 #9
    Sorry, but i did not get you in this, and how would it help me solve the question....
    the circumference i know is given by 2*pi*r so here it becomes 2*pi..
     
  11. Sep 30, 2010 #10

    ehild

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    I have no idea how can you prove this inequality if you studied only GP-s yet. In the integral method I suggested I would find a function which takes the same values as the terms of this progression at positive integers. All terms are positive, so the area under the function from x=0 to infinity is higher than the sum of the areas of the yellow rectangles, which is the same as the sum of the terms from n=1 to infinity , see the picture. But it is rather complicated. You can transform the terms in the progression so they take the form A/(1+z^2 ) and the integral of such function is A(atan(infinity) - atan (0) ) =A pi/2.


    ehild
     

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  12. Sep 30, 2010 #11
    I did get the rest..... though this part flew over my head.:confused:
    What are A and z^2 and how did the small 'a' come in the integral part?
     
  13. Oct 1, 2010 #12

    ehild

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    z is a new variable instead of n, defined as z=(n-a)b, which makes 1/(n2-n+2)=A/(1+z2). You have to find the parameters a and b.

    I meant by "atan" the function "arctangent", the inverse of tangent.

    You can not use this method without knowing calculus. I just wanted to show a possible way of solution. What did your teacher suggest, how to solve the problem?

    I made a mistake in the last formula in my previous post, as A is not the same there as it was in A/(1+z2). It is an other constant, say B.

    ehild
     
  14. Oct 1, 2010 #13
    I don't know how to use Latex in this.
    Click this link

    http://codecogs.izyba.com/gif.latex?\sum_{n%3D1}^{\infty}%20\frac{2}{2+n%28n-1%29}%20%3C%202%20\left%281+%20\sum_{n%3D2}^{\infty}%20\frac{1}{n%28n-1%29}%20\right%29%20%3D%204%3C%202%20\pi [Broken]

    Hope you got it now
     
    Last edited by a moderator: May 5, 2017
  15. Oct 1, 2010 #14
    Hey, thanks bud!!
    this is a nice way!!
     
    Last edited by a moderator: May 5, 2017
  16. Oct 1, 2010 #15
    Ok ..... i am understanding a bit now.
    Actually i'm learning these things on my own so this isn't exactly homework(no teacher help) .... just questions from a book(without solutions:wink:) or any other source.

    Abdul Quadeer's method was quite nice and it didn't have to use calculus too.
     
  17. Oct 2, 2010 #16

    ehild

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    Yes, it was an ingenious solution!

    ehild
     
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