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look at this series:

$$\sum\limits_{n = 1}^\infty {\dfrac{{{{( - 1)}^{n - 1}}}} {{2n - 1}}{x^{2n}}}$$

by Cauchy-Hadamard formula,the above series convergence region is $(-1,1)$. at the end points, it is a alternating series. so its convergence region is $[-1,1]$.

my question is: how to find a function which Taylor expansion equal the above series?

thanks very much

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Easier to take $\sum \frac{(-1)^{n-1}}{2n-1}x^{2n-1}$ and take the derivative, which is easily summed. – Thomas Andrews Dec 17 '12 at 15:43
up vote 1 down vote accepted

$$\sum\limits_{n = 1}^\infty {\dfrac{{{{( - 1)}^{n - 1}}}} {{2n - 1}}{x^{2n}}}=x\sum\limits_{n = 1}^\infty {\dfrac{{{{( - 1)}^{n - 1}}}} {{2n - 1}}{x^{2n-1}}} \overset{def}{=}xg(x).$$ Then $$g'(x)=\sum\limits_{n = 1}^\infty {{( - 1)}^{n - 1} {x^{2n-2}}}=\sum\limits_{n = 1}^\infty {{( - 1)}^{n - 1} {(x^2)^{n-1}}}=\sum\limits_{n = 1}^\infty {{( - x^2)}^{n - 1}}=\sum\limits_{k = 0}^\infty {{( - x^2)}^{k}}=\frac{1}{1+x^2}.$$

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For every $x \in [-1,1]$ we have \begin{eqnarray} f(x) &=&\sum_{n=1}^\infty\frac{(-1)^{n-1}}{2n-1}x^{2n} =x\sum_{n=0}^\infty\frac{(-1)^n} {2n+1}x^{2n+1}=x\sum_{n=0}^\infty(-1)^n\int_0^xt^{2n}\,dt\\ &=&x\int_0^x\sum_{n=0}^\infty(-1)^nt^{2n}\,dt=x\int_0^x\frac{1}{1+t^2}\,dt =x\arctan x. \end{eqnarray}

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Hint: The series $$\sum\limits_{n = 0}^\infty {\dfrac{{{{( - 1)}^{n}}}} {{2n + 1}}{x^{2n+1}}}$$ is the Taylor series of $\arctan x$ and converges for $\left|x\right|\le1$.

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