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Math Help - series with integral test

  1. #1
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    series with integral test

    \sum_{n=3}^{\infty}\frac{n^2}{e^n}

    i have to tell if this series is convergent or divergent, so i figured i could try the integral test here. the problem is, once i get to the stage of substituting with:

    \int_3^{\infty} \ \frac{x^2}{e^x}\, dx

    i get a little bit lost because integration really isn't my strong point >_<

    i realize that after i have integrated i need to put it in limit form and evaluate it, i just can't get the integration part done. please help?
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  2. #2
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    I do not understand your "integral test"

    You know that \lim_{n \rightarrow +\infty} \frac{n^4}{e^n} = 0

    Therefore there exists N such that for every n > N, \:\frac{n^4}{e^n} < 1 or \frac{n^2}{e^n} < \frac{1}{n^2}

    And you know that the series \sum_{k=1}^{n} \frac{1}{n^2} converges

    Another method is to calculate the limit of \frac{u_{n+1}}{u_n}
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  3. #3
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    i just learned the integral test yesterday, it (apparently) says that if you have a positive, decreasing, & continuous function that you can replace your sequence with, you can use the improper integral of that, turn it into a limit, and if that converges, the sequence converges. if it diverges, the sequence diverges. (or so i was told)

    edit:
    & i don't really see why this is:

    \lim_{n \rightarrow +\infty} \frac{n^4}{e^n} = 0

    Therefore there exists N such that for every n > N, \:\frac{n^4}{e^n} < 1 or \frac{n^2}{e^n} < \frac{1}{n^2}
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  4. #4
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    Quote Originally Posted by buttonbear View Post
    \sum_{n=3}^{\infty}\frac{n^2}{e^n}

    i have to tell if this series is convergent or divergent, so i figured i could try the integral test here. the problem is, once i get to the stage of substituting with:

    \int_3^{\infty} \ \frac{x^2}{e^x}\, dx
    This becomes a whole lot easier if you write it as \int_3^\infty x^2e^{-x} dx. Then integrate by parts, twice.

    i get a little bit lost because integration really isn't my strong point >_<

    i realize that after i have integrated i need to put it in limit form and evaluate it, i just can't get the integration part done. please help?
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  5. #5
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    so, trying to integrate this by parts:
    \int_3^\infty x^2e^{-x} dx

    i first said
    \ u= x^2,  dv=e^-xdx
    \ du= 2xdx, v=-e^-x

    then i got
    -e^-x*x^2- integral of -e^-x*2xdx
    so i did parts again, with

    \ u= 2x,  dv=-e^-xdx
    \ du= 2dx, v=e^-x

    =-e^-x*x^2+e^-x*2x+ integral of 2e^-xdx
    =-e^-x*x^2+e^-x*2x-2xe^-x

    i really feel like i'm doing something wrong,
    can someone help?

    p.s.- i apologize for the last few lines..LaTex was giving me some trouble >.<
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  6. #6
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    Quote Originally Posted by buttonbear View Post
    i just learned the integral test yesterday, it (apparently) says that if you have a positive, decreasing, & continuous function that you can replace your sequence with, you can use the improper integral of that, turn it into a limit, and if that converges, the sequence converges. if it diverges, the sequence diverges. (or so i was told)
    You are right, I forgot this
    I apologize
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  7. #7
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    oh, that's okay
    i'm actually not that sure of myself at all when it comes to these kinds of problems, anyway

    but, do you have any idea where i went wrong in my integration above?
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  8. #8
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    You made a mistake at the last step when integrating 2e^-x

    \int 2e^{-x}dx = -2\:e^{-x}

    Finally

    \int x^2e^{-x}dx = (-x^2-2x-2)\:e^{-x}
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  9. #9
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    oh, okay, i thought i had messed up a lot more than that >.<

    although, i'm not really sure why there's a negative sign in that middle term
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  10. #10
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    There is one

    Deriving (-x^2-2x-2)\:e^{-x} gives (-2x-2)\:e^{-x}-(-x^2-2x-2)\:e^{-x}=x^2\:e^{-x}

    While deriving (-x^2+2x-2)\:e^{-x} gives (-2x+2)\:e^{-x}-(-x^2+2x-2)\:e^{-x}=(x^2-4x+4)\:e^{-x}
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  11. #11
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    oh, wow, that is a big difference then!
    i see how it derives to a different thing, but, i don't know how to get it
    thanks though!

    also, getting back to the bigger question
    i plugged the integrated version back into the limit from 3 to infinity

    and i ended up with something like..
    (-infinity^2- 2(infinity)- 2)e^-infinity- (-3^2- 2(3)- 2)e^-2

    i figured the first part of that was going to go to zero because of the e^-infinity

    does this mean that the integral converges, & thus, the series converges?

    (sorry again, LaTex hates me)
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  12. #12
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    Taking (-x^2-2x-2)\:e^{-x} in +\infty gives 0 because \lim_{x \rightarrow +\infty}x^k e^{-x}=0 for every k real

    Therefore the integral converges to (3^2+2\cdot 3+2)\:e^{-3}=17 \:e^{-3} and thus the series converge
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  13. #13
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    okay, well i must have made a mistake in my integrating when i missed the negative sign..

    thank you so much
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  14. #14
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    \int x^2\:e^{-x} dx

    \ u= x^2,  dv=e^{-x}dx
    \ du= 2xdx, v=-e^{-x}

    \int x^2e^{-x} dx = -x^2\:e^{-x} - \int -2x\:e^{-x} dx=-x^2\:e^{-x} + \int 2x\:e^{-x} dx

    \ u= 2x,  dv=e^{-x}dx
    \ du= 2dx, v=-e^{-x}

    \int 2x\:e^{-x} dx = -2x\:e^{-x} - \int -2\:e^{-x} dx = -2x\:e^{-x} + \int 2\:e^{-x} dx = -2x\:e^{-x} - 2\:e^{-x}
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