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Math Help - Series.

  1. #1
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    Series.

    Hello,

    I have two problems with series first \sum^{\infty }_{n=1}\frac{(-1)^nn^2}{n^3+1}

    so I will like to know of this is absolute convergent here for I use two test.

    First \frac{n^2}{n^3+1} =^?0 and I think this is oké \frac{n^2(1)}{n^2(n+\frac{1}{n^2})}= \frac{1}{n+ \frac{1}{n^2}}=0

    then the second test \frac{n^2}{n^3+1}<\frac{(n+1)^2}{(n+1)^3+1} but the dominator will be bigger then te nominator?

    Or not?

    Second problem \sum^{\propto}_{n=1}\frac{n}{2n+1}

    I will proof that this is convergent \frac{\frac{(n+1)}{2n+3}}{\frac{n}{2n+1}}=\frac{n+  1}{2n+3}\frac{2n+1}{n}

    if I calculated this then I get \frac{5}{2} thus not convergent but it must be.

    Who can help me out? Greets.
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  2. #2
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    Hello, Bert!

    I have an "eyeball" approach to #1 . . . which may not be acceptable.

    1)\;\;\sum^{\infty}_{n=1}\frac{(-1)^nn^2}{n^3+1}
    Is this absolutely convergent?
    For large n:\;\;\frac{n^2}{n^3 + 1} \approx \frac{1}{n} . . . which is the divergent Harmonic Series.

    Hence the given series also diverges.


    2)\;\;\sum^{\infty}_{n=1}\frac{n}{2n+1}
    \lim_{n\to\infty} a_n\;=\;\lim_{n\to\infty}\frac{n}{2n+1}

    Divide top and bottom by n\;\;\lim_{n\to\infty}\frac{1}{2 + \frac{1}{n}} \;= \;\frac{1}{2 + 0} \;= \;\frac{1}{2}

    Since the n^{th} term does not approach 0, the series diverges.
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  3. #3
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    but normally you can test so harmonica serie with two test or not?
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  4. #4
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    For the first one, You can show absolute divergence by using the integral test.

    Consider:

    \int_1^\infty \frac{x^2}{x^3+1}\,dx

    Integrate by substitution to get:

    \lim_{b\to\infty} \left( \frac{1}{3}\ln\left|b^3+1\right| - \frac{1}{3}\ln 2 \right) = \infty

    Since this integral diverges, so does \sum_{n=1}^\infty \frac{n^2}{n^3+1} (which is the absolute value of the terms of the alternating sum).
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