# Math Help - How does cos/sin affect convergence of series

1. ## How does cos/sin affect convergence of series

{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

2. Originally Posted by sj247

{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
Use Dirichlet's test to prove that both these sums converge. Note that since $\displaystyle \sum \cos(r)$ is bounded and $\displaystyle \frac{1}{r^2+1}\to 0$ for example.

3. Originally Posted by sj247

{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...

$\displaystyle \frac{|\cos (r+1)|}{r^{2}+1} < \frac{1}{r^{2}}$ (1)

... and ...

$\displaystyle \frac{|\sin (r^{2}+1)|}{r^{2}+1} < \frac{1}{r^{2}}$ (2)

... and the series $\displaystyle \sum_{n=1}^{\infty} \frac{1}{r^{2}}$ converges, also the two series You proposed converge...

Kind regards

$\chi$ $\sigma$

4. Originally Posted by chisigma
Because is...

$\displaystyle \frac{|\cos (r+1)|}{r^{2}+1} < \frac{1}{r^{2}}$ (1)

... and ...

$\displaystyle \frac{|\sin (r^{2}+1)|}{r^{2}+1} < \frac{1}{r^{2}}$ (2)

... and the series $\displaystyle \sum_{n=1}^{\infty} \frac{1}{r^{2}}$ converges, also the two series You proposed converge...

Kind regards

$\chi$ $\sigma$
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 :|

5. 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.

6. Originally Posted by Drexel28
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 :|

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

7. Originally Posted by tonio
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.

8. Originally Posted by sj247
...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...
We now consider the case ot the sums $\displaystyle \sum_{k=1}^{n} \sin k$ and $\displaystyle \sum_{k=1}^{n} \cos k$ and that is valid for all the cases...

$\displaystyle \sum_{k=0}^{n-1} a\ r^{k} = a\ \frac{1-r^{n}}{1-r}$ (1)

... and setting $r=a= e^{i}$ we obtain in some steps...

$\displaystyle S= \sum_{k=1}^{n} e^{i k} = e^{i}\ \frac{1-e^{i n}}{1-e^{i}}$

$\displaystyle = \frac{e^{i}}{2} \ \frac{1- \cos 1 - \cos n + \cos (n-1) + i\ \{\sin 1 - \sin n + \sin (n-1) \}} {1 - \cos 1}$ (2)

Without proceeding further it is clear from (2) that both the quantities...

$\displaystyle \sum_{k=1}^{n} \cos k = \mathcal{R} \{S\}$

$\displaystyle \sum_{k=1}^{n} \sin k = \mathcal{I} \{S\}$

... are bounded...

Kind regards

$\chi$ $\sigma$