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

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

    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
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by sj247 View Post
    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
    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.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by sj247 View Post
    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 \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
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chisigma View Post
    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 :|
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  5. #5
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    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.
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  6. #6
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    Quote Originally Posted by Drexel28 View Post
    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
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by tonio View Post
    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.
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  8. #8
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by sj247 View Post
    ...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...

    We start with the 'geometric sum'...

    \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
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