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Math Help - Evaluating Series.

  1. #1
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    Evaluating Series.

    Find the sum of each of the following series,

    (i) \sum^\infty_{n=1} nx^n
    (ii) \sum^\infty_{n=1} \frac{n}{2^n}
    (iii) \sum^\infty_{n=1} n(n-1)x^n

    Use Taylor series to evaluate the following limits:
    (a) \lim_{x \rightarrow 0} \frac{1 - \frac{1}{2}x^2 - cos x}{x^4}
    (b) \lim_{x \rightarrow 0} \frac{x - arctan x}{x^3}
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by panda* View Post
    Find the sum of each of the following series,

    (i) \sum^\infty_{n=1} nx^n
    (ii) \sum^\infty_{n=1} \frac{n}{2^n}
    (iii) \sum^\infty_{n=1} n(n-1)x^n

    Use Taylor series to evaluate the following limits:
    (a) \lim_{x \rightarrow 0} \frac{1 - \frac{1}{2}x^2 - cos x}{x^4}
    (b) \lim_{x \rightarrow 0} \frac{x - arctan x}{x^3}
    1. Consider \sum_{n=0}^{\infty}x^n. Because its interval of convergence is (-1,1) on that same interval it is uniformly convergent. Also note that \forall{x}\in(-1,1)~\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}. So now since we know this series is uniformly convergent we may say this \begin{aligned}\left(\frac{1}{1-x}\right)'&=\frac{d}{dx}\sum_{n=0}^{\infty}x^n\\<br />
&=\sum_{n=1}^{\infty}nx^{n-1}\\<br />
&\implies{x}\cdot\left(\frac{1}{1-x}\right)'=\sum_{n=0}^{\infty}nx^n~\forall{x}\in(-1,1)<br />
\end{aligned}


    2. Direct application of number one


    3. Using the same agrument as before we can see that

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

    3. Using the everywhere convergent Maclaurin series we may rewrite our limit as follows

    \begin{aligned}\lim_{x\to{0}}\frac{1-\frac{x^2}{2}-\cos(x)}{x^4}&=\frac{1-\frac{x^2}{2}-\left(1-\frac{x^2}{2}+\frac{x^4}{24}-\cdots\right)}{x^4}\\<br />
&=\lim_{x\to{0}}\frac{\frac{-x^4}{24}+\cdots}{x^4}\\<br />
&=\frac{-1}{24}<br />
\end{aligned}

    4. We know that \forall{x}\in(-1,1]~\arctan(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{2n+1}. So since if S=\left\{x:x\to{0}\right\} we can see that S\subset(-1,1]. Therefore we may rewrite our limit as

    \begin{aligned}\lim_{x\to{0}}\frac{x-\arctan(x)}{x^3}&=\lim_{x\to{0}}\frac{x-\left(x-\frac{x^3}{3}+\cdots\right)}{x^3}\\<br />
&=\lim_{x\to{0}}\frac{\frac{x^3}{3}-\cdots}{x^3}\\<br />
&=\frac{1}{3}<br />
\end{aligned}
    Last edited by Mathstud28; November 14th 2008 at 06:31 PM. Reason: typo
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    [snip]

    3. Using the everywhere convergent Maclaurin series we may rewrite our limit as follows

    \begin{aligned}\lim_{x\to{0}}\frac{1-\frac{x^2}{2}-\cos(x)}{x^4}&=\frac{1-\frac{x^2}{2}-\left(1-\frac{x^2}{{\color{red}2}}+\frac{x^4}{24}-\cdots\right)}{x^4}\\<br />
&=\lim_{x\to{0}}\frac{\frac{-x^4}{24}+\cdots}{x^4}\\<br />
&=\frac{-1}{24}<br />
\end{aligned}

    [snip]
    Not trying to be super picky, but shouldn't that be a two? Otherwise, the limit would diverge...

    --Chris
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    Not trying to be super picky, but shouldn't that be a two? Otherwise, the limit would diverge...

    --Chris
    Yes, it was a typo. Thank you.
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