-- Series Convergence

[rancid moth]

I thought i would try my luck posting this one again with the new filter:

I am trying to discuss the convergence of the following series, but I'm
getting nowhere fast:

S = sum(n=0,oo) (4*n+1) / [( b + 2*n*(b+c) )*n!*Gamma(1/2 - n) ] * z^(2*n)
*P(2*n,cos(Y)) for 0<=Y<=pi

where P(n,x) is the legendre polynomial, b and c are >0 and real. my guess
is that |z| must be <1, but does this need to hold for all Y? my particular
interest is about the point |z|=1.

my attempt would be the following...

Consider the same series of S however with absolute values. Since
|P(n,x)|<=1, for -1<=x<=1 each term in the series is then less than or equal
to

a_n = |(4*n+1) / [( b + 2*n*(b+c) )*n!*Gamma(1/2 - n) ] |*| z^(2*n)| .

The ratio test then shows that this is convergent for |z|<1, and so the
series is absolutely convergent there.

Then I consider |a_n+1/a_n| for |z|=1, and prove the bound |a_n+1/a_n| >
(4n+5)*(1+2n)*n / (2*(4n+1)*(1+n)^2). I can then solve this recurrence
relation and get |a_n| > 2*(1+4n)Gamma(1/2+n)/(5*n^2*Sqrt(pi)*Gamma(n)) *
|a_1|. Now Gamma(1/2+n)/Gamma(n) = sqrt(n)-8/sqrt(n) +O(n^-3/2) as n->oo.
and so the terms of the series behave as 8/(5*sqrt(pi*n)) + O(n^-3/2) and so
by the limit comparison theorem, the series diverges and therefore by
comparison the series of |a_n| diverges. Therefore the series is not
bsolutely convergent for |z|=1. But is it convergent? and if so, for what
values of Y?

cheers
moth