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Math Help - Convergence of a Series

  1. #1
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    Question Convergence of a Series

    Can anyone help me with the following problem please:

    Using convergence tests, determine whether the series

    ∑ (-1)n/√(n+5) over n=1 and n=

    converges or diverges.

    Many thanks to anyone who can provide me with any advice on solving this problem.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi

    \sum_{n=1}^{\infty}(-1)^n\frac{1}{\sqrt{n+5}}

    This series is alternating hence you can try to show the convergence using the alternating series test.
    Last edited by flyingsquirrel; May 14th 2008 at 11:58 AM. Reason: typo spotted by Mathstud :)
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by flyingsquirrel View Post
    Hi

    \sum_{k=1}^{\infty}(-1)^n\frac{1}{\sqrt{n+5}}

    This series is alternating hence you can try to show the convergence using the alternating series test.
    Be careful with that n not k...

    \sum_{n=0}^{\infty}\frac{(-1)^{n}}{\sqrt{n+5}}

    Expanding upon Flying squirrels explanation...since a_{n+1}<a_n

    Where a_n=\frac{1}{\sqrt{n+5}}

    and also since \lim_{n\to\infty}a_n=\lim_{n\to\infty}\frac{1}{\sq  rt{n+5}}=0 we can see by the alternating series test that this series is at least conditionally convergent
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Mathstud28 View Post
    Be careful with that n not k...

    \sum_{n=0}^{\infty}\frac{(-1)^{n}}{\sqrt{n+5}}

    Expanding upon Flying squirrels explanation...since a_{n+1}<a_n

    Where a_n=\frac{1}{\sqrt{n+5}}

    and also since \lim_{n\to\infty}a_n=\lim_{n\to\infty}\frac{1}{\sq  rt{n+5}}=0 we can see by the alternating series test that this series is at least conditionally convergent
    Alternating series test: If the limit of the absolute value of the terms is 0 and beyond some point the sequence of absolute values is non-increasing [that is there exists an N such that for all n>N; a_n>=a_(n+1)] then the series converges (conditionally).

    RonL
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Alternating series test: If the limit of the absolute value of the terms is 0 and beyond some point the sequence of absolute values is non-increasing [that is there exists an N such that for all n>N; a_n>=a_(n+1)] then the series converges (conditionally).

    RonL
    How is this different than what I said? I didnt state some of the things but what I said was correct
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by Mathstud28 View Post
    How is this different than what I said? I didnt state some of the things but what I said was correct
    You have not shown convergence untill you have shown that the series is non-increasing beyound some point, and you have not stated that this is needed. Nor have you done the alternative, which is to just say that this is needed but obvious.

    RonL
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    You have not shown convergence untill you have shown that the series is non-increasing beyound some point, and you have not stated that this is needed. Nor have you done the alternative, which is to just say that this is needed but obvious.

    RonL
    So you are saying that a_{n+1}<a_n,\forall{n}>0 Isnt enough?
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by Mathstud28 View Post
    So you are saying that a_{n+1}<a_n,\forall{n}>0 Isnt enough?
    That is not what you had in your post you had that the \lim_{n \to \infty}|a_n|=0

    But neither of the conditions:

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

    or

    \exists N;\ \forall n>N,\ |a_{n+1}| \le |a_n|

    on its own is enough you need both.

    RonL
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Be careful with that n not k...

    \sum_{n=0}^{\infty}\frac{(-1)^{n}}{\sqrt{n+5}}

    Expanding upon Flying squirrels explanation...since a_{n+1}<a_n

    Where a_n=\frac{1}{\sqrt{n+5}}

    and also since \lim_{n\to\infty}a_n=\lim_{n\to\infty}\frac{1}{\sq  rt{n+5}}=0 we can see by the alternating series test that this series is at least conditionally convergent
    if you look closely I had both....I didnt state that n>0 but I had that condition

    and I took \lim_{n\to\infty}\frac{1}{\sqrt{n+5}}=\lim_{n\to\i  nfty}\bigg|\frac{(-1)^{n}}{\sqrt{n+5}}\bigg|

    are you saying that my prolem is that I just did it and did not explicity state what I was doing?
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  10. #10
    Grand Panjandrum
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    Quote Originally Posted by Mathstud28 View Post
    if you look closely I had both....I didnt state that n>0 but I had that condition

    and I took \lim_{n\to\infty}\frac{1}{\sqrt{n+5}}=\lim_{n\to\i  nfty}\bigg|\frac{(-1)^{n}}{\sqrt{n+5}}\bigg|

    are you saying that my prolem is that I just did it and did not explicity state what I was doing?
    OK

    RonL
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