# Proof of a inequality

## Homework Statement

Prove that

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

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.

## Answers and Replies

Related Precalculus Mathematics Homework Help News on Phys.org
please give me a starting tip atleast !!!

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

ehild

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

ehild

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.

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
Homework Helper
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

#### Attachments

• 6.3 KB Views: 274
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
Homework Helper
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

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

Last edited by a moderator:
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
Homework Helper
Abdul Quadeer's method was quite nice and it didn't have to use calculus too.
Yes, it was an ingenious solution!

ehild