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Math Help - Sums using the integral test

  1. #1
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    Sums using the integral test

    sum e^n/(1+e^2n)
    n = 1 to infinity

    By using the integral test I need to integrate this to determine whether it will converge or diverge.

    Can someone please help with the integration. I cannot figure out how to integrate it.
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by ur5pointos2slo View Post
    sum e^n/(1+e^2n)
    n = 1 to infinity

    By using the integral test I need to integrate this to determine whether it will converge or diverge.

    Can someone please help with the integration. I cannot figure out how to integrate it.
    \int \frac{e^n}{1+e^{2n}} dn

    use substution,

    let u = e^{n}, then du = e^n dn \implies dn = \frac{du}{e^n}

    Do the rest! show where you are getting stuck
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    Quote Originally Posted by harish21 View Post
    \int \frac{e^n}{1+e^{2n}} dn

    use substution,

    let u = e^{n}, then du = e^n dn \implies dn = \frac{du}{e^n}

    Do the rest! show where you are getting stuck
    My problem is just that. The derivative of e^n is just e^n dn. What kind of technique would I need to use on this?
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    Quote Originally Posted by ur5pointos2slo View Post
    My problem is just that. The derivative of e^n is just e^n dn. What kind of technique would I need to use on this?
    scratch that stupid me. Ok we are using a u-substitution. I am confused on where to go next. There must be a trick using exponential rules?
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  5. #5
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by ur5pointos2slo View Post
    My problem is just that. The derivative of e^n is just e^n dn. What kind of technique would I need to use on this?
    du = e^n dn \implies dn = \frac{du}{e^n}

     \therefore \int \frac{e^n}{1+e^{2n}} dn

    = \int \frac{1}{1+u^2} du = tan^{-1}(u)

    plug back u = e^n, you have:

    [tan^{-1}(e^n)]_1^\infty

    now find out if the series converges or not!
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  6. #6
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    Quote Originally Posted by harish21 View Post
    du = e^n dn \implies dn = \frac{du}{e^n}

     \therefore \int \frac{e^n}{1+e^{2n}} dn

    = \int \frac{1}{1+u^2} du = tan^{-1}(u)

    plug back u = e^n, you have:

    [tan^{-1}(e^n)]_1^\infty

    now find out if the series converges or not!
    That seemed too easy. I couldn't see it until you pointed it out. So it equals Pi/4 and it converges.
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  7. #7
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by ur5pointos2slo View Post
    That seemed too easy. I couldn't see it until you pointed it out. So it equals Pi/4 and it converges.
    \frac{\pi}{4}??

    [tan^{-1}(e^n)]_1^\infty = (tan^{-1}(e^{\infty})) - (tan^{-1}(e^1)) = (tan^{-1}(\infty)) - (tan^{-1}(e)) = \frac{\pi}{2} - tan^{-1}(e)  , This is not equal to \frac{\pi}{4}
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  8. #8
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    Quote Originally Posted by harish21 View Post
    \frac{\pi}{4}??

    [tan^{-1}(e^n)]_1^\infty = (tan^{-1}(e^{\infty})) - (tan^{-1}(e^1)) = (tan^{-1}(\infty)) - (tan^{-1}(e)) = \frac{\pi}{2} - tan^{-1}(e)
    oops forgot the exponential...so Pi/2 - tan^-1(e) converges.
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  9. #9
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by ur5pointos2slo View Post
    oops forgot the exponential...so Pi/2 - tan^-1(e) converges.
    Yes!
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