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Math Help - Working with Power & Taylor series. Interval of convergence. General help.

  1. #1
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    Working with Power & Taylor series. Interval of convergence. General help.

    I'm having trouble with the algebraic manipulation required to find the interval of convergence of many of the Power series and Taylor series problems in my calculus class.

    For example:

    \sum_{n=0}^{\infty}\left ( -1 \right )^{n}\frac{n}{4^{n}}x^{2n}


    Using the Ratio test, and ignoring \left ( -1 \right )^{n} because we can, we have:
    \left | \frac{a_{n+1}}{a_{n}} \right|=\frac{(n+1)x^{(2n+1)}}{4^{n+1}}*\frac{4^{n  }}{nx^{2n}}

    The 4^{n} can be cancelled:

    \frac{(n+1)x^{(2n)}*x}{4}*\frac{1}{nx^{2n}}

    Distributing the x^{2n}:

    \frac{nx^{(2n)}+x^{(2n)}*x}{4}*\frac{1}{nx^{2n}}

    Cancelling the nx^{2n} terms:

    =\frac{1+x^{(2n)}*x}{4}*\frac{1}{1}=\frac{1+x^{(2n  +1)}}{4}

    The next step:

    \lim_{n \to \infty }\frac{1+x^{(2n+1)}}{4}

    I want to say that the limit is infinity but I think that would mean the interval of convergence is all real numbers; I don't think that is the answer. Perhaps I made algebraic/arithmetic mistake or perhaps I am not considering the limit correctly.
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  2. #2
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    Re: Working with Power & Taylor series. Interval of convergence. General help.

    Remember that multiplication distributes over a term in parentheses. (n+1)x^{2(n+1)} = (n+1)x^{2n+2} = (n+1)x^{2n}x^2 = nx^{2n}x^2 + x^{2n}x^2

    So, you can cancel x^{2n} in the numerator and denominator (leave the n for now).

    \begin{align*}\lim_{n\to \infty} \left|\dfrac{a_{n+1}}{a_n}\right| & = \lim_{n\to \infty} \dfrac{(n+1)x^2}{4n} \\ & = \lim_{n \to \infty}\dfrac{nx^2 + x^2}{4n} \\ & =  \lim_{n \to \infty} \left(\dfrac{nx^2}{4n} + \dfrac{x^2}{4n}\right) \\ & = \dfrac{x^2}{4}\end{align*}
    Thanks from jamesrb
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