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Math Help - Testing an Infinite Sum of a Series Made up of a Difference for Convergence

  1. #1
    Junior Member Coop's Avatar
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    Testing an Infinite Sum of a Series Made up of a Difference for Convergence

    Hi,

    I am asked to determine if the infinite sum of [1/5n - 1/(5n+3)] is convergent.

    I tested each term individually and found that they both diverge, which means that I cannot say whether their difference converges or diverges.

    Can anyone tell me what I should do when I encounter a difference or sum of series and they both diverge individually?

    I found that the series above is greater than [1/5n^2 - 1(5n+3)] and that the first term in this converges while the second diverges still. So wouldn't that mean that their difference diverges, and that the original series diverges by the Comparison Test?

    WolframAlpha says it converges ???

    infinite sum (1/(5n)-1/(5n+3)) - Wolfram|Alpha

    Any help? What should I do?

    Thanks
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  2. #2
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    Re: Testing an Infinite Sum of a Series Made up of a Difference for Convergence

    Hint: try to put the terms together into one fraction. Then you can apply this general rule:

    If you are dealing with the ratio of two polynomials, the series will converge if and only if the highest power of the denominator exceeds the highest power of the numerator by more than 1.

    For example,

    \sum_{n=1}^{\infty} \frac{n^2+10000000n}{n^4} will converge

    On the other hand,

    \sum_{n=1}^{\infty} \frac{n^3+10n}{100000n^4}

    Will diverge.

    See if you can apply the rule in this situation.
    Last edited by SworD; April 4th 2013 at 01:28 PM.
    Thanks from Coop
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  3. #3
    Junior Member Coop's Avatar
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    Re: Testing an Infinite Sum of a Series Made up of a Difference for Convergence

    Quote Originally Posted by SworD View Post
    Hint: try to put the terms together into one fraction. Then you can apply this general rule:

    If you are dealing with the ratio of two polynomials, the series will converge if and only if the highest power of the denominator exceeds the highest power of the numerator by more than 1.

    For example,

    \sum_{n=1}^{\infty} \frac{n^2+10000000n}{n^4} will converge

    On the other hand,

    \sum_{n=1}^{\infty} \frac{n^3+10n}{100000n^4}

    Will diverge.

    See if you can apply the rule in this situation.
    Okay thanks I will try that. But can you say that the infinite sum of a difference is divergent if one of the individual sums is convergent and the other is divergent?
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  4. #4
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    Re: Testing an Infinite Sum of a Series Made up of a Difference for Convergence

    Quote Originally Posted by Coop;780093
    I am asked to determine if the infinite sum of [1/5n - 1/(5n+3)
    is convergent.

    This is really just a simple comparison test.

    \sum\limits_{n = 1}^\infty  {\left( {\frac{1}{{5n}} - \frac{1}{{5n + 3}}} \right)}  = \sum\limits_{n = 1}^\infty  {\left( {\frac{3}{{25{n^2} + 15n}}} \right)}  \leqslant \frac{3}{{25}}\sum\limits_{n = 1}^\infty  {\left( {\frac{1}{{{n^2}}}} \right)}

    Can you see the comparison?
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  5. #5
    Junior Member Coop's Avatar
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    Re: Testing an Infinite Sum of a Series Made up of a Difference for Convergence

    Quote Originally Posted by Plato View Post
    This is really just a simple comparison test.

    \sum\limits_{n = 1}^\infty  {\left( {\frac{1}{{5n}} - \frac{1}{{5n + 3}}} \right)}  = \sum\limits_{n = 1}^\infty  {\left( {\frac{3}{{25{n^2} + 15n}}} \right)}  \leqslant \frac{3}{{25}}\sum\limits_{n = 1}^\infty  {\left( {\frac{1}{{{n^2}}}} \right)}

    Can you see the comparison?
    Right thanks, I think that's where SworD was going. I see the comparison but I am having a little trouble seeing your first equality. :/

    P.S. Is my statement above your post true? About the infinite sum of a difference converging.
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    Re: Testing an Infinite Sum of a Series Made up of a Difference for Convergence

    Quote Originally Posted by Coop View Post
    I see the comparison but I am having a little trouble seeing your first equality.

    Are you saying that you have trouble adding two fractions?
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  7. #7
    Junior Member Coop's Avatar
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    Re: Testing an Infinite Sum of a Series Made up of a Difference for Convergence

    Nvm I figured it out, thanks for the help SworD and Plato.
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    Re: Testing an Infinite Sum of a Series Made up of a Difference for Convergence

    Quote Originally Posted by Coop View Post
    But can you say that the infinite sum of a difference is divergent if one of the individual sums is convergent and the other is divergent?
    This is true.
    Quote Originally Posted by Coop View Post
    I found that the series above is greater than [1/5n^2 - 1(5n+3)] and that the first term in this converges while the second diverges still. So wouldn't that mean that their difference diverges, and that the original series diverges by the Comparison Test?
    It is also true that \sum_{n=1}^\infty(1/(5n^2) - 1/(5n+3)) diverges and that 1/(5n) - 1/(5n+3)\ge1/(5n^2) - 1/(5n+3). However, it is not true that |1/(5n) - 1(5n+3)|\ge|1/(5n^2) - 1/(5n+3)|. If the latter inequality were true, then the original series would diverge by the comparison test.
    Thanks from Coop
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  9. #9
    Junior Member Coop's Avatar
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    Re: Testing an Infinite Sum of a Series Made up of a Difference for Convergence

    Quote Originally Posted by emakarov View Post
    This is true.
    It is also true that \sum_{n=1}^\infty(1/(5n^2) - 1/(5n+3)) diverges and that 1/(5n) - 1/(5n+3)\ge1/(5n^2) - 1/(5n+3). However, it is not true that |1/(5n) - 1(5n+3)|\ge|1/(5n^2) - 1/(5n+3)|. If the latter inequality were true, then the original series would diverge by the comparison test.
    Thank you
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