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Math Help - Alternating Series

  1. #1
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    Alternating Series

    I have trouble with alternating series.
    I have to test the series for convergence or divergence.
    Any help would be appreciated
    Thanks!
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    Last edited by BACONATOR; July 18th 2008 at 10:04 AM. Reason: forgot to write down the problem.
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  2. #2
    Moo
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    Quote Originally Posted by BACONATOR View Post
    I have trouble with alternating series.
    I have to test the series for convergence or divergence.
    Any help would be appreciated
    Thanks!
    Hello,

    Remember that for an alternating series to converge \sum_{n} (-1)^n \ a_n, you have to check that :

    - \lim_{n \to \infty} a_n=0
    - a_n is a decreasing sequence



    But here, one can see that when n approaches high values, +2 can be neglected.

    Thus the general term is equivalent to (-1)^n \frac{n}{\sqrt{n^3+2}} \sim (-1)^n \frac{n}{\sqrt{n^3}}=(-1)^n \frac{n}{n^{3/2}}=(-1)^n \frac{1}{n^{1/2}}, which is an alternating Riemann series that converges.
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  3. #3
    Super Member flyingsquirrel's Avatar
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    Hi
    Quote Originally Posted by Moo View Post
    Thus the general term is equivalent to (-1)^n \frac{n}{\sqrt{n^3+2}} \sim (-1)^n \frac{n}{\sqrt{n^3}}=(-1)^n \frac{n}{n^{3/2}}=(-1)^n \frac{1}{n^{1/2}}, which is an alternating Riemann series that converges.
    You can't use approximations ( \sim) in this case since the sign of (-1)^n\frac{n}{\sqrt{n^3+2}} changes : The theorem you're using states that if {\color{red}a_n,\,b_n>0}, if a_n\sim b_n and if \sum a_n converges then \sum b_n converges.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by BACONATOR View Post
    I have trouble with alternating series.
    I have to test the series for convergence or divergence.
    Any help would be appreciated
    Thanks!
    We can see that this satisfies the two conditions of the converging series

    \sum_{n=0}^{\infty}(-1)^na_n

    Firstly

    \exists{N}\backepsilon\forall{n>N}\quad{a_{n+1}<a_  n}

    Furthermore

    \lim_{n\to\infty}a_n=0

    \Rightarrow\quad\sum_{n=0}^{\infty}(-1)^n\frac{n}{\sqrt{n^3+2}}~~\text{converges}
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