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Math Help - convergence/divergence series advice

  1. #1
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    convergence/divergence series advice

    I've almost done my latest assignment, but for a couple questions of which I'm hopelessly stuck at. I'm not sure how to proceed with them.
    If anyone could give me advice on how to get started on them, I'll be a very happy crustacean.
    1. Prove:
    \lim_{n \to \infty} \frac {n^p}{a^n} =0 where p>0 & a>1

    2. Test for convergence/divergence:
    \sum_{n=1}^{\infty} \frac {(n+1)}{n^3 ln(n+2)}
    what test do I use here? Integral? If yes, what should I make u=?

    3. For what values of k isthe following series absolutely convergent? For what values k>=0 conditionally convergent?
    \sum_{n=3}^{\infty} \frac {(-1)^n}{n*lnn(ln(ln n))^k}
    where do I even begin here?! I hate logs!

    Final Question:
    a_{n}>0 for all n, and \frac{a_{n+1}}{a_{n}} \to L>0
    Show that the power series:
    \sum a_{n} (x - x_{0})^n has a radius of convergence R=\frac{1}{L}
    This is what I did:
    Use Ratio test:
    \frac {a_{n+1} (x - x_{0})^{n+1}}{a_{n} (x - x_{0})^n}

    \frac{a_{n+1}}{a_{n}} (x - x_{0})

    since we know \frac{a_{n+1}}{a_{n}}\to L

    L(x - x_{0}) < 1

    (x - x_{0}) < \frac{1}{L}

    thus R=\frac{1}{L}
    Is this the correct way to prove this? I ask because it just seems too easy. I'm always of the opinion that if I find it easy it's likely because I've done something wrong!

    Thanks for any and all help.
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  2. #2
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    Quote Originally Posted by Dr Zoidburg View Post
    3. For what values of k isthe following series absolutely convergent? For what values k>=0 conditionally convergent?
    \sum_{n=3}^{\infty} \frac {(-1)^n}{n*lnn(ln(ln n))^k}
    where do I even begin here?! I hate logs!
    This is so hard to read.

    Is it \sum_{n = 3}^{\infty}\frac{(-1)^n}{n(\ln{n})[\ln{(\ln{n})}]^k}?

    Please use spacing or brackets to make things easier to read...
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  3. #3
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    Quote Originally Posted by Prove It View Post
    This is so hard to read.

    Is it \sum_{n = 3}^{\infty}\frac{(-1)^n}{n(\ln{n})[\ln{(\ln{n})}]^k}?

    Please use spacing or brackets to make things easier to read...
    Yes it is. Sorry about the lousy setting out. It's what's on the assignment.
    Anyhoo....any suggestions/advice to get me started on this beast?
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  4. #4
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    I've been trying to do these but have had scant luck thus far.
    This is where I've got up to:
    1. Prove \lim_{n \to \infty} \frac {n^p}{a^n} =0where p>0 & a>1
    I've tried the Ratio test, but I got this:
    \frac {(n+1)^p}{a^{(n+1)}}. \frac {a^n}{n^p}

    = \frac{1}{a}. \frac {(n+1)^p}{n^p}

    As \lim_{n \to \infty} \frac {(n+1)^p}{n^p} = 1, leaving me with = \frac{1}{a} which obviously isn't 0.
    What am I doing wrong here?


    As for this one:
    \sum_{n = 3}^{\infty}\frac{(-1)^n}{n(\ln{n})[\ln{(\ln{n})}]^k}
    I figure that under absolute convergence, (-1)^n = 1. So that gets rid of that part. To simplify the bottom, I make u=ln(n) & du = 1/n dn.
    Where do I go from there?
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