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Math Help - Find the radius of convergence and the interval of conv...

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    Find the radius of convergence and the interval of conv...

    convergence, also indicating the type of convergence, either absolutely or conditionally.

    Σ [n=2 to n=infinity] [(-1)^n (2x+3)^n]/(nln(n))

    I tried to take the absolute root test to cancel the nth powers, but I am stumped by the natural log. Can someone please tell me where to start?



    Last edited by hbarnes; July 18th 2013 at 07:21 PM.
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    Re: Find the radius of convergence and the interval of conv...

    Have you tried the ratio test?
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    Re: Find the radius of convergence and the interval of conv...

    Hello, hbarnes!

    Find the interval of converence.
    Also, indicate the type of convergence: absolute or conditional.

    . . \displaystyle \sum^{\infty}_{n=2} \frac{(-1)^n(2x+3)^n}{n\ln(n)}

    I tried to take the absolute root test to cancel the nth powers,
    but I am stumped by the natural log.

    Ratio Test

    R \:=\:\left|\frac{a_{n+1}}{a_n}\right| \:=\:\left|\frac{(2x+3)^{n+1}}{(n+1)\ln(n+1)}\cdot  \frac{n\ln(n)}{(2x+3)^n}\right|

    . . . . . . . . . =\;\left|\frac{(2x+3)^{n+1}}{(2x+3)^n}\cdot\frac{n  }{n+1}\cdot\frac{\ln(n)}{\ln(n+1)}\right|

    . . . . . . . . . =\;\left|(2x+3)\cdot\frac{\frac{1}{n}}{\frac{1}{n+  1}} \cdot\frac{\ln(n)}{\ln(n+1)}\right|

    \displaystyle\lim_{n\to\infty}R \;=\;\lim_{n\to\infty}|2x+3|\cdot\lim_{n\to\infty}  \left|\frac{\frac{1}{n}}{\frac{1}{n+1}}\right| \cdot \lim_{n\to\infty}\left|\frac{\ln(n)}{\ln(n+1)} \right|

    \displaystyle\lim_{n\to\infty}R \;=\;|2x+3|\cdot 1 \cdot \lim_{n\to\infty}\left|\frac{\ln(n)}{\ln(n+1)} \right| . [1]


    For that last limit, apply L'Hopital:

    . . \lim_{n\to\infty}\frac{\frac{1}{n}}{\frac{1}{n+1}} \;=\;\lim_{n\to\infty}\frac{n+1}{n} \;=\;\lim_{n\to\infty}\frac{1+\frac{1}{n}}{1} \;=\;1

    Hence, [1] becomes: . |2x+3|\cdot 1\cdot 1 \:=\:|2x+3|


    We have:

    . . \begin{array}{c}\qquad\;\;\;|2x+3| \:<\:1 \\ \\ -1 \:<\:2x+3\:<\:1 \\ \\ -4\:<\:2x\:<\:-2 \\ \\ -2\:<\:x\:<\:-1 \end{array}


    The series is absolutely convergent.
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    Re: Find the radius of convergence and the interval of conv...

    That may not be the complete interval of convergence, as we don't know anything about the convergence of the series when the ratio of terms is 1.

    So substitute x = -2 and x = -1 into your function and use a different test to determine the convergence of those two series.
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