Prove the sum of this series =pi/3

[santa]

prove the sum of this series =pi/3

$$1 +\frac{ 1}{5} -\frac{ 1}{7} - \frac{1}{11 }+\frac{1}{13} + \frac{1}{17} - ...$$

 

[CompuChip]

Surely you must have some idea.
Why don't you start by writing it out in sum notation ($$\sum_{n = 0}^\infty \cdots$$)?

 

[Gib Z]

Cheeky CompuChip :P You know he knows the solution, he just wants to tease us :P Look at the rest of his posts to see what I mean.

I've got a nice solution for this though =] I'll post it when someone else asks for it, because I want to let some other guys give this a try. It's not actually that hard, this one.

 

[sadhu]

a twisted Gregorian pi series , missing fractions of odd multiples of 3, is all what it is.

Cheeky CompuChip :P You know he knows the solution, he just wants to tease us :P Look at the rest of his posts to see what I mean.

i Am not sure whether he know all the solutions. all his posts where only questions no replies or answers.
even if he knows all the answer , he should be interested in knowing different method of doing it.which usually people do less they are descendent's of Newton or Einstein .

 

[Xalos]

Pi happens to be a transcendal no. meaning-you can't get its value by any algebraic method. If this a puzzle, you could get a nobel prize for the answer.

 

[Moo Of Doom]

Pi happens to be a transcendal no. meaning-you can't get its value by any algebraic method. If this a puzzle, you could get a nobel prize for the answer.

That's not what transcendental means. Transcendental just means that it is not the root (zero) of any polynomial with integer coefficients. This has nothing to do with infinite sums such as this one, and there are many such sums (and products as well) that are quite well-known. The Wikipedia article on Pi and this MathWorld article list quite a lot of them.

 

[CompuChip]

If this a puzzle, you could get a nobel prize for the answer.

A Nobel Prize in what? Mathematics?

 

[Gib Z]

Not to mention, The Nobel Prize is not award to Mathematicians unless the work has had a serious application in one of the awarded fields, such as the case of John Nash. I doubt this one would.

Anyone made any progress? It sounds like some of you already have a solution, just can't be bothered to post it >.<" Ill post mine tomorrow.

EDIT: CompuChip beat me to it !

 

[CompuChip]

Hmm, can't be that hard if even I solved it in a little under 5 minutes. (Though I must admit, having the answer already helped me think of arctan a lot sooner )

 

[sadhu]

believe me it won't take more than a five min or even less than a min

 

[Gib Z]

Aww don't say that! I wanted to sound really smart for having a solution! Ruin my fun :(

 

[santa]

thanks for all

from these http://mathworld.wolfram.com/PiFormulas.html


let$$S=1- \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - ... = \frac{\pi}{4}.$$

thus $$\frac{1}{3} S= \frac{1}{3} - \frac{1}{9}+ \frac{1}{15} - \frac{1}{21} + ... = \frac{\pi}{12}.$$

series = $$S + \frac{1}{3} S =\frac{\pi}{3}$$

 

[Xalos]

Sorry for my answer! I am just a grade10 student and I just had read of the word in the library. I will remember not to reply about thing I don't know. Sorry Gib Z and thanks Moo of Doom!

 

[BryanP]

Heh, interesting.

 

[sadhu]

that mine method too