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Thread: Alternating Series - Converges or Diverges

  1. #1
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    Alternating Series - Converges or Diverges

    Question:

    Test the series for convergence or divergence.
    $\displaystyle
    sn = \sum (-1)^n \frac{2n-1}{5n+1}
    $

    My attempt:

    This is an alternating series, therefore, I will use the alternating series test.

    We know that sn+1 < sn, however, the limit of sn is $\displaystyle \frac{2}{5}$ and not 0. Therefore, this series diverges.

    Am I correct?

    Edit: I tried the ration test: $\displaystyle \frac{sn+1}{sn}$ and it showed that the series is convergent because $\displaystyle \lim \frac{sn+1}{sn} = 0
    $
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by calc101 View Post
    Question:

    Test the series for convergence or divergence.
    $\displaystyle
    s_n = \sum (-1)^n \frac{2n-1}{5n+1}
    $

    My attempt:

    This is an alternating series, therefore, I will use the alternating series test.

    We know that sn+1 < sn, however, the limit of sn is $\displaystyle \frac{2}{5}$ and not 0. Therefore, this series diverges.

    Am I correct?
    That's correct!
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by calc101 View Post
    Edit: I tried the ration test: $\displaystyle \frac{sn+1}{sn}$ and it showed that the series is convergent because $\displaystyle \lim \frac{sn+1}{sn} = 0
    $
    When you use the ratio test for absolute convergence, you have

    $\displaystyle \lim\left|\left(-1\right)^n\frac{2n+1}{5n+6}\cdot\frac{5n+1}{2n-1}\right|=\lim\frac{\left(2n+1\right)\left(5n+1\ri ght)}{\left(5n+6\right)\left(2n-1\right)}=1$ which shows that the test is inconclusive.
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  4. #4
    Super Member Showcase_22's Avatar
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    We know that sn+1 < sn
    When $\displaystyle n=1$, $\displaystyle s_1=-\frac{1}{6}$ but $\displaystyle s_{n+1}=s_2=-\frac{1}{6}+\frac{3}{11}$. Clearly, in this case, $\displaystyle s_{n+1}> s_n$.

    Did you mean "$\displaystyle a_n$ is a decreasing sequence where $\displaystyle a_n= \left|(-1)^n \frac{2n-1}{5n+1} \right|=\frac{2n-1}{5n+1}$"?
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  5. #5
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    Quote Originally Posted by Showcase_22 View Post
    When $\displaystyle n=1$, $\displaystyle s_1=-\frac{1}{6}$ but $\displaystyle s_{n+1}=s_2=-\frac{1}{6}+\frac{3}{11}$. Clearly, in this case, $\displaystyle s_{n+1}> s_n$.

    Did you mean "$\displaystyle a_n$ is a decreasing sequence where $\displaystyle a_n= \left|(-1)^n \frac{2n-1}{5n+1} \right|=\frac{2n-1}{5n+1}$"?
    I am sorry, I meant the sequence, an. Thank you for correcting me.
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  6. #6
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    since the limit is 2/5 and not equal to zero, then the series is divergent.
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  7. #7
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    Quote Originally Posted by dr.tea View Post
    since the limit is 2/5 and not equal to zero, then the series is divergent.
    Well of course $\displaystyle \lim _{n \to \infty } \left( { - 1} \right)^n \frac{{2n - 1}}{{5n + 1}} \ne \frac{2}{5}$.

    Here is the first test we all should apply.
    The series $\displaystyle \sum {a_n } $ converges only if $\displaystyle \left( {a_n } \right) \to 0$.

    So answer this question does $\displaystyle \left( { - 1} \right)^n \frac{{2n - 1}}{{5n + 1}} \to 0 ?$

    If not then $\displaystyle \sum {\left( { - 1} \right)^n \frac{{2n - 1}}{{5n + 1}}} $ diverges.
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  8. #8
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    The test of alternating series is
    $\displaystyle \sum(-1)^{n}a_{n}$ converges if
    1) $\displaystyle a_{n}$ is decreasing
    2) \the limit of $\displaystyle a_{n}$ =0

    since the second condition is not satisfied , then the sequence diverges
    ( lim $\displaystyle a_{n}$=2/5 )
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