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Math Help - Series

  1. #1
    Super Member 11rdc11's Avatar
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    Series

    Ok so I'm working on this problem but have problem understanding it. I got the right answer but I made an assumption to get there. I need help proving that assumption.

    Here is the problem

    Find the values of p for which the series in convergent

    \sum_{n=1}^{\infty} n(1+n^2)^p

    Ok so I used the integral test because f(x) = x(1+ x^2)^p is continous, positive, and decreasing(this is where I made the assumption that it was decreasing).

    \int_{1}^{\infty} x(1+x^2)^{p}dx = \lim_{t \to \infty}\bigg[\frac{(1+x^2)^{p+1}}{2p+2}\bigg]_1^t = \lim_{t \to \infty}\bigg(\frac{1}{2}\bigg)\bigg(\frac{(1+t^2)^  {p+1}}{p+1}\bigg) - \frac{2^p}{p+1}

    Thus the limit exist when p + 1 < 0 so the series converges when p< -1.

    Is that right? Also how do I show that the function is decreasing(the assumption I need to prove). I took the 1st derivative and equalled it to less than 0 to show me where it was decreasing but ran into problems.

    This is what I get stuck on with the derivative

    x^2(1+2p)< -1

    Thanks
    Last edited by 11rdc11; September 3rd 2008 at 01:49 PM.
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  2. #2
    Super Member 11rdc11's Avatar
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    Can I say the function is decreasing since when p is negative the denominator is growing faster than the numerator so it is decreasing?
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  3. #3
    Moo
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    Hello,

    Yes, it is continuous.

    Actually, my reasoning is as follows :
    The series {a_n} converges ---> lim a_n = 0

    By taking the contrapositive of this assumption, it's evident to say that lim a_n has to be 0.

    Assuming p<0 (otherwise, one can see it is not possible), let q=-p for more convenience :

    \frac{n}{(1+n^2)^q} \to_{\substack{n \to \infty}} \frac{n}{n^{2q}}

    ---> for it to tend to 0, q has to be > 1/2, which is p<-1/2. Necessary condition, not sufficient.

    -----------------------------------------------
    If you want to show it's decreasing :

    x(1+x)^p

    Its derivative is (1+x^2)^p+p 2x^2(1+x^2)^{p-1}=(1+x^2)^{p-1} (1+x^2+2px^2)

    (1+x^2)^{p-1} \ge 0 because x \ge 1

    1+(2p+1)x^2
    We want to find p such that it is \le 0

    if 2p+1\ge 0 ---> (2p+1)x+1 is always positive. Which is not possible.
    if 2p+1<0, since x>1, (2p+1) has to be < -1, that is to say p<-1


    etc.
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