Proof of a inequality

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Homework Statement



Prove that

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


Homework Equations


none


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:
 

Answers and Replies

  • #2
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please give me a starting tip atleast !!! :approve:
 
  • #5
ehild
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I do not have any other idea. What have you learnt about infinite series?

ehild
 
  • #7
ehild
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Try to approximate the circumference of the unit circle by inscribed polygons.

ehild
 
  • #8
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I do not have any other idea. What have you learnt about infinite series?

ehild
all i know is about diverging and converging GP (infinite) and sum of decreasing infinite GP.
 
  • #9
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Try to approximate the circumference of the unit circle by inscribed polygons.

ehild
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..
 
  • #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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  • #11
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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
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?
 
  • #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
 
  • #13
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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:
  • #14
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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
Hey, thanks bud!!
this is a nice way!!
 
Last edited by a moderator:
  • #15
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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
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.
 
  • #16
ehild
Homework Helper
15,526
1,901
Abdul Quadeer's method was quite nice and it didn't have to use calculus too.
Yes, it was an ingenious solution!

ehild
 

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