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Math Help - converge or divergent arctan series

  1. #1
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    converge or divergent arctan series

    heres the problem: http://img208.imageshack.us/img208/6449/23355228cp8.png

    its known that arctan x is (-pi/2, pi/2) and its domain is (-infinity, infinity)

    so the limit is lim x-> infinity arctan x = pi/2

    would that be enough to show that the series is convergent? or must i use a series test.

    also, could i show this using the squeeze theorem also? between the pi/2's

    thanks alot.
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    Quote Originally Posted by rcmango View Post
    heres the problem: http://img208.imageshack.us/img208/6449/23355228cp8.png

    its known that arctan x is (-pi/2, pi/2) and its domain is (-infinity, infinity)

    so the limit is lim x-> infinity arctan x = pi/2

    would that be enough to show that the series is convergent? or must i use a series test.

    also, could i show this using the squeeze theorem also? between the pi/2's

    thanks alot.
    you must use a series test. so far, the techniques you have hinted at using work for SEQUENCES (taking the limit of the nth term formula to see if it converges, using the Squeeze theorem etc), not series. (in fact, if you want to use limits to tell whether a series DOES NOT converge, there is the divergence test, but it is not applicable here)

    the series diverges by the Limit Comparison test. Compare to the harmonic series
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    Okay, still need help. I've done a few things to get closer to the solution but i don't think i'm quite right.

    alright i was looking over the limit comparison test, So, An = arctan/n

    Can An also = (pi/2) / n ?

    Or should i just forget about squeeze altogether with series.

    Bn = 1/n (a harmonic series).

    I know using this test, I divide An by Bn, I get pi/2 for a finite C. which the harmonic series (Bn) does diverge, thus An diverges.

    The arctan is messing me all up.

    I also thought using the direct comparison test was easier? where the arctan equation is bigger than the harmonic series, so it diverges by the rules.

    thanks for the help.
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    Quote Originally Posted by rcmango View Post
    Okay, still need help. I've done a few things to get closer to the solution but i don't think i'm quite right.

    alright i was looking over the limit comparison test, So, An = arctan/n

    Can An also = (pi/2) / n ?
    doing that makes no sense. because \frac {\arctan n}n \le \frac {\frac {\pi}2}n for all n \in \mathbb{N}. since the harmonic series diverges, we can't draw any conclusion from this

    Or should i just forget about squeeze altogether with series.
    that's a good idea, at least for questions like these

    Bn = 1/n (a harmonic series).

    I know using this test, I divide An by Bn, I get pi/2 for a finite C. which the harmonic series (Bn) does diverge, thus An diverges.
    yes

    The arctan is messing me all up.
    you seem to work with it fine so far

    I also thought using the direct comparison test was easier?
    what would you compare it with?

    where the arctan equation is bigger than the harmonic series, so it diverges by the rules.
    this is not true. look at n = 1. arctan(n)/n is not bigger than 1/n for ALL n = {1,2,3, ...}
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    I was taking any number and plugging it into for n, for example 2.

    so arctan 2 / 2 and then 1/2 so i assumed .5535

    was bigger than 1/2 and it behaved this way for many bigger numbers. So i believed the arctan to be bigger. I guess i can't not assume this.


    If i can't use the direct comparison easily. I need help to set up the limited comparison test.

    I'm not sure if i need to change the arctan to something or just leave it as it is, and divide through by the harmonic series?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by rcmango View Post
    I was taking any number and plugging it into for n, for example 2.

    so arctan 2 / 2 and then 1/2 so i assumed .5535

    was bigger than 1/2 and it behaved this way for many bigger numbers. So i believed the arctan to be bigger. I guess i can't not assume this.


    If i can't use the direct comparison easily. I need help to set up the limited comparison test.
    ok

    I'm not sure if i need to change the arctan to something or just leave it as it is, and divide through by the harmonic series?
    leave it as is and divide through by 1/n. have you looked up the test in your text book?
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  7. #7
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    Yes, have looked through it.

    No examples for arctan.

    so we have (arctan * n) / n * (n/1)

    = arctan/n + 1 ?

    so just 1 is the limit?

    and since C = 1.

    they both diverge because were comparing the the harmonic series?
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