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Find the sum of power series $$\sum_{n=1}^{\infty}\frac{(-1)^nx^{n-1}}{2n-1}$$

How can we use derivation/integration method for this series?

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$$\sum_{n=1}^{\infty}\frac{(-1)^nx^{n-1}}{2n-1}$$ is convergent for $-1<x<1$. Then, in case of $0<x<1$ and with $X=x^{1/2}$ Let: $$y(X)=\sum_{n=1}^{\infty}\frac{(-1)^nX^{2n-1}}{2n-1}$$ (wich is different from the above series. Do not confuse them. We will come back to the above series at the end). $$y'(X)=\sum_{n=1}^{\infty}(-1)^nX^{2n-2}=-\frac{1}{1+X^2} \qquad \text{geometric series}$$

$$Y(X)=-\int \frac{dx}{1+X^2}=-\tan^{-1}(X)+c$$ $X=0 \quad\to\quad Y(0)=\sum_{n=1}^{\infty}\frac{(-1)^n 0^{2n-1}}{2n-1}=0=-\tan^{-1}(0)+c \quad\to\quad c=0$ $$y(X)=\sum_{n=1}^{\infty}\frac{(-1)^nX^{2n-1}}{2n-1}=-\tan^{-1}(X)$$ With $X=x^{1/2}$ $$\sum_{n=1}^{\infty}\frac{(-1)^nx^{n-1/2}}{2n-1}=-\tan^{-1}(x^{1/2})$$ $$x^{-1/2}\sum_{n=1}^{\infty}\frac{(-1)^nx^{n}}{2n-1}=\tan^{-1}(x^{1/2})$$ $$\sum_{n=1}^{\infty}\frac{(-1)^nx^{n}}{2n-1}=-x^{1/2}\tan^{-1}(x^{1/2})$$ Do similar calculus in case of $-1<x<0$.

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  • $\begingroup$ If you don't want the OP to be confused do not use $\;X\;$ in your second-third line. Use $\;a,\,\alpha,\,\omega\;$ or something like that. There's no problem at all to differentiate/integrate wrt $\;a,\,\alpha,\,\omega\;$ orwhatever. $\endgroup$ – DonAntonio Feb 11 '17 at 11:12
  • $\begingroup$ @JJacquelin, Why did we chose substitution $x^{1/2}$? $\endgroup$ – user300045 Feb 11 '17 at 12:54
  • $\begingroup$ The denominator is $2n-1$. If we want to cancel it through differentiation, the power of $X$ must be $2n-1$. Since the actual power is $n-1$ we have to change $x^n$ into $X^{2n}$, so $x$ into $X^2$. $\endgroup$ – JJacquelin Feb 11 '17 at 14:05
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For $x\ge0,$

$$\dfrac{(-1)^nx^{n-1}}{2n-1}=\dfrac ix\dfrac{(i\sqrt x)^{2n-1}}{2n-1}$$

Now $$\ln(1+y)-\ln(1-y)=\sum_{r=1}^\infty\dfrac{y^{2r-1}}{2r-1}$$

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  • $\begingroup$ Could you show more intuitive method? $\endgroup$ – user300045 Feb 11 '17 at 10:41
  • $\begingroup$ @user_99, Not sure about intuitive term! Whenever the $n$ term of contains a linear expression of $n$ in the denominator , Log Series comes to my mind. Whenever the $n$ term of contains a factorial of $n$ in the denominator, e series. Whenever there is an Arithmetic Series in the numerator=> Binomial Series $\endgroup$ – lab bhattacharjee Feb 11 '17 at 10:51
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{n = 1}^{\infty}{\pars{-1}^{n}\,x^{n - 1} \over 2n - 1} & = \sum_{n = 1}^{\infty}\pars{-1}^{n}x^{n - 1}\int_{0}^{1}y^{2n - 2}\,\dd y = -\int_{0}^{1}\sum_{n = 1}^{\infty}\pars{-xy^{2}}^{n - 1}\,\dd y = -\int_{0}^{1}{\dd y \over 1 + xy^{2}} \\[5mm] & = -\,{1 \over \root{x}}\int_{0}^{\root{x}}{\dd y \over y^{2} + 1} = \bbx{\ds{-\,{\arctan\pars{\root{x}} \over \root{x}}}} \end{align}

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