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Math Help - Integral Test

  1. #1
    Newbie Foxtenn5's Avatar
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    Integral Test

    Use the integral test to determine if the given series is convergent or divergent.

    the sum from k=1 to infinity of e^k / (2+e^k)


    math, math, math...

    the limit as b approaches infinity of [lnb-ln1]

    -ln1 = 0

    the limit approaches 0, so the series is convergent.


    Does this look correct?
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  2. #2
    Moo
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    Hello,

    \lim_{b \to +\infty} \ln(b)=+\infty \neq 0
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  3. #3
    Super Member flyingsquirrel's Avatar
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    Wooo, how did you get that [\ln b-\ln 1] ? Plus, unfortunately, \lim_{b\to\infty}\ln b= \infty

    What is \int_1^b\frac{\exp x}{2+\exp x}\,\mathrm{d}x ?
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  4. #4
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    \int_{1}^{\infty}\frac{e^{k}}{2+e^{k}}dk={\infty}

    I am pretty sure this one is divergent.
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  5. #5
    Newbie Foxtenn5's Avatar
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    Ok, let's see...

    limit as b approaches infinity (from 1 to b) of e^x/(2+e^x)

    u = 2 + e^x
    du = e^x

    so, by substitution method,

    limit as b approaches infinity (from 1 to b) of 1/u du

    limit as b approaches infinity (from 1 to b) of[lnu + c]

    limit as b approaches infinity of [lnb - ln1]

    so, since ln(infinity) = infinity and ln1 = 0, does that mean that the series diverges?

    ...or am I totally off?
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  6. #6
    Moo
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    Quote Originally Posted by Foxtenn5 View Post
    Ok, let's see...

    limit as b approaches infinity (from 1 to b) of e^x/(2+e^x)

    u = 2 + e^x
    du = e^x

    so, by substitution method,

    limit as b approaches infinity (from 1 to b) of 1/u du

    limit as b approaches infinity (from 1 to b) of[lnu + c]

    limit as b approaches infinity of [lnb - ln1]

    so, since ln(infinity) = infinity and ln1 = 0, does that mean that the series diverges?

    ...or am I totally off?
    This is it
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