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Math Help - Convergence interval

  1. #1
    Bop
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    Convergence interval

    Hello, I have this serie:

    \displaystyle\sum_{i=1}^\infty{(\displaystyle\frac  {1}{n}-tg(\displaystyle\frac{1}{n}))}<br />

    I have tried to apply ratio test but it is indefinite. Any ideas?

    Thank you!
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Bop View Post
    Hello, I have this serie:

    \displaystyle\sum_{i=1}^\infty{(\displaystyle\frac  {1}{n}-tg(\displaystyle\frac{1}{n}))}<br />

    I have tried to apply ratio test but it is indefinite. Any ideas?

    Thank you!
    Is this asking to determine the convergence/divergence for \sum_{n\in\mathbb{N}}\left\{\frac{1}{n}-\tan\left(\frac{1}{n}\right)\right\}. Note that \tan\left(\frac{1}{n}\right)=\frac{1}{n}+\frac{1}{  n^3}+\cdots. Thus, we can see that \frac{1}{n}-\tan\left(\frac{1}{n}\right)\underset{n\to\infty}{  \sim}\frac{-1}{3n^3}. So, your series converges.
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  3. #3
    Bop
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    How do you know that

    tan(\frac{1}{x})=\frac{1}{x}+\frac{1}{x^{3}}+.. ?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Bop View Post
    How do you know that

    tan(\frac{1}{x})=\frac{1}{x}+\frac{1}{x^{3}}+.. ?
    It should be \frac{1}{x}+\frac{1}{3x^3}+\cdots. It's the Maclaurin series for \tan(x).

    Alternatively, suppose that \frac{1}{n}-\tan\left(\frac{1}{n}\right) shared convergence with some series \frac{1}{n^\lambda}. Then, we get that \lim_{\infty}\frac{\frac{1}{n}-\tan\left(\frac{1}{n}\right)}{n^\lambda}=C for some constant C. Letting z=\frac{1}{n} gives \lim_{z\to0}\frac{z-\tan(z)}{z^\lambda}=C. Try to find a \lambda that works.
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  5. #5
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    Quote Originally Posted by Drexel28 View Post
    It should be \frac{1}{x}+\frac{1}{3x^3}+\cdots. It's the Maclaurin series for \tan(x).

    Alternatively, suppose that \frac{1}{n}-\tan\left(\frac{1}{n}\right) shared convergence with some series \frac{1}{n^\lambda}. Then, we get that \lim_{\infty}\frac{\frac{1}{n}-\tan\left(\frac{1}{n}\right)}{n^\lambda}=C for some constant C. Letting z=\frac{1}{n} gives \lim_{z\to0}\frac{z-\tan(z)}{{\color{red} z^\lambda}}=C. Try to find a \lambda that works.
    I think it is typo, Right?
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by TWiX View Post
    I think it is typo, Right?
    No. that part is correct. It should have been \frac{\frac{1}{n}-\tan\left(\frac{1}{n}\right)}{\frac{1}{n^\lambda}} originally.
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  7. #7
    Bop
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    With \lambda=2 the limits is \frac{1}{2}, so it converges.

    Thank you very much!
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  8. #8
    Bop
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    Hi, I was thinking, I thought we could only apply this method with positive terms series, and this has negative terms, so, is it right?
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  9. #9
    Bop
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    Hi, I have taken a look at my book and it says this test is for postive series, is it wrong?
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