Maybe I'm being pedantic, but I don't like to use that here. The fact that the absolute convergence of a series implies the convergence of the series is only true since the real (or complex numbers) is a complete metric space and since this seems like a calculus question I'm not sure if that's ok. I don't know if Dirichlet's test is much easier though :|
So in my first example {sum 1, inf} cos(r+1)/r^2+1
(An)=1/(r^2 +1)
(Bn)= cos(r+1)
i have to show (An) is decreasing & {sum}(Bn) is bounded ?
Not sure how to go about showing {sum}(Bn) is bounded. Although obviously it must as (Bn) must be between 1, -1 for any n.
I'm not sure of the above: Dirichlet's test's proof does use completeness of the complex (real) field in a very
clear and decisive way (at least the proofs I know of), whereas absolute convergence ==> convergence is true
without using (at least directly, in one of the proofs I know of) completeness...
Tonio
I was just making a remark. There are proofs of Dirichlet's theorem which don't appeal directly to completeness. That said, I do agree that the most common one (the one using summation by parts) does. I guess it's all kind of irrelevant since everything (and I mean this in a very, very loose sense) depends upon completeness if one goes far enough back.
We now consider the case ot the sums and and that is valid for all the cases...
We start with the 'geometric sum'...
(1)
... and setting we obtain in some steps...
(2)
Without proceeding further it is clear from (2) that both the quantities...
... are bounded...
Kind regards