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$$\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)(2n+2)3^{n+1}}$$ Since $$\tan^{-1}x=\int \frac{1}{1+x^{2}} dx=\int (1-x^{2}+x^{4}+...)dx=x-\frac{x ^3}{3}+\frac{x^5}{5}+...$$ so$$\int \tan^{-1}x dx=\int (x-\frac{x^3}{3}+\frac{x^5}{5}+...)dx=\frac {x^2}{2}-\frac{x^4}{3\cdot4}+\frac {x^6}{5\cdot 6}+...=\sum_{n=0}^{\infty} \frac{(-1)^nx^{2n+2}}{(2n+1)(2n+2)}$$ Therefore, $$\int_{0}^{1/{\sqrt 3}}\tan^{-1}xdx=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)(2n+2)3^{n+1}}$$ $$\Rightarrow \frac{\tan^{-1}(\frac{1}{\sqrt 3})}{\sqrt 3}-\frac{\ln(1+\frac{1}{3})}{2}=\frac{\pi}{6\sqrt3}-\frac{\ln\frac{4}{3}}{2}$$ Is my method correct?

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  • $\begingroup$ That should be $\ln \frac{4}{3}$, not $\ln \frac{3}{4}$. $\endgroup$ – Robert Israel Mar 25 '19 at 16:37
  • $\begingroup$ Thank you. I'm just not pretty sure that whether there is any mistake. $\endgroup$ – Maggie Mar 25 '19 at 16:39
  • $\begingroup$ It's correct but you might want to show how you got $\int \arctan(x)dx$. $\endgroup$ – marty cohen Mar 25 '19 at 22:56
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You series equals $$ \frac{1}{3}\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)3^n}-\sum_{n\geq 0}\frac{(-1)^n}{(2n+2)3^{n+1}} $$ or $$ \frac{1}{\sqrt{3}}\arctan\frac{1}{\sqrt{3}}+\frac{1}{2}\sum_{n\geq 1}\frac{(-1)^n}{n 3^n}=\frac{\pi}{6\sqrt{3}}-\frac{1}{2}\log\left(1+\frac{1}{3}\right). $$ No integrals, just partial fraction decomposition, reindexing and the Maclaurin series of $\arctan(x)$ and $\log(1+x)$.

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  • $\begingroup$ Thank you. This is awesome!! $\endgroup$ – Maggie Mar 25 '19 at 16:56
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over \pars{2n + 1}\pars{2n + 2}3^{n + 1}}} = {1 \over 3}\sum_{n = 0}^{\infty}{\pars{-1/3}^{n} \over 2n + 1} - {1 \over 3}\sum_{n = 0}^{\infty} {\pars{-1/3}^{n} \over 2n + 2} \\[5mm] = &\ {1 \over 3}\sum_{n = 0}^{\infty}{\pars{-1/3}^{n} \over 2n + 1} + \sum_{n = 1}^{\infty}{\pars{-1/3}^{n} \over 2n} \\[5mm] = &\ {1 \over 3}\,{\root{3} \over \ic}\sum_{n = 0}^{\infty}{\pars{\ic/\root{3}}^{2n + 1} \over 2n + 1} + \sum_{n = 1}^{\infty}{\pars{\ic/\root{3}}^{2n} \over 2n} \\[5mm] = &\ {1 \over 3}\,{\root{3} \over \ic}\sum_{n = 1}^{\infty}{\pars{\ic/\root{3}}^{n} \over n}\,{1 - \pars{-1}^{n} \over 2} + \sum_{n = 1}^{\infty}{\pars{\ic/\root{3}}^{n} \over n} \,{1 + \pars{-1}^{n} \over 2} \\[5mm] = &\ {\root{3} \over 3}\,\Im\sum_{n = 1}^{\infty}{\pars{\ic/\root{3}}^{n} \over n} + \Re\sum_{n = 1}^{\infty}{\pars{\ic/\root{3}}^{n} \over n} \end{align}

Since $\ds{\sum_{n = 1}^{\infty}{\pars{\ic/\root{3}}^{n} \over n} = -\ln\pars{1 - {\root{3} \over 3}\,\ic} = -\,{1 \over 2}\,\ln\pars{4 \over 3} + {\pi \over 6}\,\ic}$:

\begin{align} &\bbox[10px,#ffd]{\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over \pars{2n + 1}\pars{2n + 2}3^{n + 1}}} = {\root{3} \over 3}\,{\pi \over 6} - {1 \over 2}\ln\pars{4 \over 3} \\[5mm] = &\ \bbx{{\root{3} \over 18}\,\pi - {1 \over 2}\ln\pars{4 \over 3}} \approx 0.1585 \end{align}

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