Can You Prove This Infinite Series Inequality?

[The legend]

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

$$T(n) = \frac{2}{2 + n(n-1)}$$

any help appreciated.

 

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please give me a starting tip atleast !

 

[ehild]

Think of the integral test of convergence.

http://en.wikipedia.org/wiki/Integral_test_for_convergence

ehild

 

[The legend]

Think of the integral test of convergence.

http://en.wikipedia.org/wiki/Integral_test_for_convergence

ehild

Is there any other way instead of using calculus. Since i don't know much of it..(in process of learning ... don't know much integration yet!)

 

[ehild]

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

ehild

 

[fzero]

If you've covered any series expansions of $$\pi$$, you might be able to use the comparison test.

http://en.wikipedia.org/wiki/Comparison_test

 

[ehild]

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

ehild

 

[The legend]

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

ehild

all i know is about diverging and converging GP (infinite) and sum of decreasing infinite GP.

 

[The legend]

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..

 

[ehild]

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

 

[The legend]

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.
What are A and z^2 and how did the small 'a' come in the integral part?

 

[ehild]

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

 

[zorro]

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

Hope you got it now

 

[The legend]

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

Hope you got it now

Hey, thanks bud!
this is a nice way!

 

[The legend]

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) or any other source.

Abdul Quadeer's method was quite nice and it didn't have to use calculus too.

 

[ehild]

Abdul Quadeer's method was quite nice and it didn't have to use calculus too.

Yes, it was an ingenious solution!

ehild