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Math Help - converge/diverge

  1. #1
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    converge/diverge

    sum n=1 oo (arctan(n))^2 / 1+n^2

    how do i solve it using integral test?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jeph View Post
    sum n=1 oo (arctan(n))^2 / 1+n^2

    how do i solve it using integral test?
    Here
    Attached Thumbnails Attached Thumbnails converge/diverge-inttest.gif  
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  3. #3
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    Quote Originally Posted by jeph View Post
    sum n=1 oo (arctan(n))^2 / 1+n^2

    how do i solve it using integral test?
    Less formally than Jhevon, as \arctan(n) by definition is giving a value in the range (-\pi/2, \pi/2) \arctan(n) is bounded (in fact not only is it bounded but as n \to \infty \arctan(n) \to \pi/2). Also for large n, 1/(1+n^2) behaves like 1/n^2 so the series is absolutly convergent.

    RonL
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    Quote Originally Posted by CaptainBlack View Post
    Less formally than Jhevon, as \arctan(n) by definition is giving a value in the range (-\pi/2, \pi/2) \arctan(n) is bounded (in fact not only is it bounded but as n \to \infty \arctan(n) \to \pi/2). Also for large n, 1/(1+n^2) behaves like 1/n^2 so the series is absolutly convergent.

    RonL
    which test is being used right here? o.o
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jeph View Post
    which test is being used right here? o.o
    the comparison test. he wasn't being very formal though, since i already answered the question.

    The comparison test says:

    Let \sum a_{n} be a series where  a_{n} \geq 0 for all  n.

    (i) If \sum a_{n} converges and | b_{n}| \leq a_{n} for all n, then \sum b_{n} converges

    (ii) If \sum a_{n} = + \infty and b_{n} \geq a_{n} for all n, then \sum b_{n} = + \infty.


    CaptainBlack used the fact that:

    \sum \left| \frac {\arctan^2 n}{1 + n^2} \right| \leq \sum \frac {(\frac {\pi}{2})^2}{1 + n^2} = \frac {\pi^2}{4} \sum \frac {1}{1 + n^2} and \sum \frac {1}{1 + n^2} converges by the comparison test, if we take \sum a_{n} to be \sum \frac {1}{n^2} and \sum b_{n} to be \sum \frac {1}{1 + n^2}

    EDIT: Bare with me, i'm just learning LaTex, it will take me a while to get used to it
    EDIT 2: Finally! It is ready!
    EDIT 3: I'm so proud of myself ...even though typing that took forever. i guess i'll get faster with practice though, and after i memorize the commands
    Last edited by Jhevon; May 22nd 2007 at 04:26 PM.
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  6. #6
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    Quote Originally Posted by Jhevon View Post
    i guess i'll get faster with practice though, and after i memorize the commands
    It takes just seconds for me. As fast as I type.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    It takes just seconds for me. As fast as I type.
    my typing speed is ok...not great, but ok. i am slow because i don't know the commands. i have a LaTex tutorial up while i'm typing and i have to be refering to it constantly. i'll get a lot quicker when i know the commands
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by jeph View Post
    which test is being used right here? o.o
    None explicitly, but knowledge of the behaviour of series. This is an informal
    argument of the form that we actually use if we want to know if this thing
    converges, only later will a formal proof be constructed to make the argument rigorous.

    The purpose of such an argument is to explain why this converges, which is not always
    clear from a formal proof (Sound of Bourbaki turning in his grave from off stage right).

    Jhevon has explained what the idea is in more detail

    RonL
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