Series convergence problem

1. Dec 27, 2008

mmzaj

what are the conditions on $$\Gamma_{2n+1}$$ so as for the series to converge ?

$$m\pi=\sum^{\infty}_{n=0} \frac{(-1^{n}) \Gamma_{2n+1}}{(2n+1) r^{2n+1}}$$

m = 0,1,2,....

r is real number

is there an explicit expression for $$\Gamma_{2n+1}$$ in terms of n ?

2. Dec 27, 2008

mathman

The gamma function for positive integer arguments is essentially the same as factoral, i.e. gamma(n+1)=n!.

3. Dec 27, 2008

Mute

$\Gamma_{2n+1}$ need not have anything to do with the gamma function! It's just a coefficient labelled with a capital gamma.

As for the form of gamma... it will obviously depend on m, and it will have to depend on r, too. I'm guessing that you'll need to add more conditions on the form you want, otherwise you may not get a unique answer. (And I don't know if you can get an answer independent of r).

(Following formulae lifted from wikipedia. check external reference to ensure correctness).

For instance,

$$\frac{\pi}{4} = \sum_{n=0}^\infty = \frac{(-1)^n}{2n+1}$$,

so we could to choose $\Gamma_{2n+1} = 4 m r^{2n+1}$.

Another series is

$$\frac{\pi}{2} = \sum_{n=0}^\infty \frac{n!}{(2n+1)!!}$$

(where the (2n+1)!! = 1*3*5*...(2n+1) ), so we could choose $\Gamma_{2n+1} = 2(-1)^n m r^{2n+1} (2n+1)n!/(2n+1)!!$. (Admittedly, though, the factor of (-1)^n in there sort of breaks with the subscript notation of the coefficient).

As for general conditions on Gamma, I guess I would run through the convergence tests for series, like the ratio test. http://en.wikipedia.org/wiki/Ratio_test

Last edited: Dec 27, 2008
4. Dec 27, 2008

jostpuur

5. Dec 31, 2008

mmzaj

i think i sorted it out .

$$tan^{-1}(\frac{s}{r})=\sum^{\infty}_{n=0} \frac{(-1^{n}) s^{2n+1} }{(2n+1) r^{2n+1}}$$

where

$$\frac{s}{r}<1$$

=> $$\Gamma_{2n+1} = s^{2n+1}$$