If for example i had:
{sum 1, inf} cos(r+1)/r^2+1
or
{sum 1, inf} sin(r^2 +1)/r^2+1
Do i try get rid of the cos/sine? if so how
Or is there a general rule im missing?
Thanks
Because is...
$\displaystyle \displaystyle \frac{|\cos (r+1)|}{r^{2}+1} < \frac{1}{r^{2}}$ (1)
... and ...
$\displaystyle \displaystyle \frac{|\sin (r^{2}+1)|}{r^{2}+1} < \frac{1}{r^{2}}$ (2)
... and the series $\displaystyle \displaystyle \sum_{n=1}^{\infty} \frac{1}{r^{2}}$ converges, also the two series You proposed converge...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
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 $\displaystyle \displaystyle \sum_{k=1}^{n} \sin k$ and $\displaystyle \displaystyle \sum_{k=1}^{n} \cos k$ and that is valid for all the cases...
We start with the 'geometric sum'...
$\displaystyle \displaystyle \sum_{k=0}^{n-1} a\ r^{k} = a\ \frac{1-r^{n}}{1-r}$ (1)
... and setting $\displaystyle r=a= e^{i}$ we obtain in some steps...
$\displaystyle \displaystyle S= \sum_{k=1}^{n} e^{i k} = e^{i}\ \frac{1-e^{i n}}{1-e^{i}} $
$\displaystyle \displaystyle = \frac{e^{i}}{2} \ \frac{1- \cos 1 - \cos n + \cos (n-1) + i\ \{\sin 1 - \sin n + \sin (n-1) \}} {1 - \cos 1} $ (2)
Without proceeding further it is clear from (2) that both the quantities...
$\displaystyle \displaystyle \sum_{k=1}^{n} \cos k = \mathcal{R} \{S\} $
$\displaystyle \displaystyle \sum_{k=1}^{n} \sin k = \mathcal{I} \{S\} $
... are bounded...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$