# Homework Help: Proof of a inequality

1. Sep 29, 2010

### The legend

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

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

any help appreciated.

2. Sep 30, 2010

### The legend

please give me a starting tip atleast !!!

3. Sep 30, 2010

### ehild

4. Sep 30, 2010

### The legend

5. Sep 30, 2010

### ehild

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

ehild

6. Sep 30, 2010

### fzero

7. Sep 30, 2010

### ehild

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

ehild

8. Sep 30, 2010

### The legend

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

9. Sep 30, 2010

### The legend

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. Sep 30, 2010

### 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

#### Attached Files:

• ###### progression.JPG
File size:
6.3 KB
Views:
95
11. Sep 30, 2010

### The legend

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?

12. Oct 1, 2010

### 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

13. Oct 1, 2010

### zorro

I don't know how to use Latex in this.

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
14. Oct 1, 2010

### The legend

Hey, thanks bud!!
this is a nice way!!

Last edited by a moderator: May 5, 2017
15. Oct 1, 2010

### The legend

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.

16. Oct 2, 2010

### ehild

Yes, it was an ingenious solution!

ehild