Trying to find an explicit sum of an infinite series: $\sum_{n=1}^{\infty} \frac{ (-1)^n }{(2n+1) 3^n }$

Question

I have the following series

$$\sum_{n=1}^{\infty}\frac{(-1)^n }{(2n+1)3^n}$$

I am trying to find an explicit sum. I know this looks like $\arctan x = \sum_{n=0}^{\infty} \frac{ (-1)^n x^{2n+1}}{2n+1} $. I do the following

$$ \sum_{n=1}^{\infty} \frac{ (-1)^n }{(2n+1) 3^n } = \sum_{n=0}^{\infty} \frac{ (-1)^n }{(2n+1) 3^n } - 1 =\sqrt{3} \sum_{n=1}^{\infty} \frac{ (-1)^n \sqrt{1/3}^{2n+1}}{(2n+1) } - 1$$

Since $(1/3)^n = (\sqrt{1/3})^2n = \sqrt{3} (\sqrt{1/3})^{2n+1}$. Thus, the sum is $$ \sqrt{3} \sum_{n=1}^{\infty} \frac{ (-1)^n \sqrt{1/3}^{2n+1}}{(2n+1) } - 1 = \sqrt{3} \arctan(1/\sqrt{3}) = 1 = \boxed{\frac{ \sqrt{3} \pi }{6} - 1 }$$

Is this a correct solution? Do you guys a differenti method?

Answer 1

As $\log\dfrac{1+x}{1-x}=2\sum_{r=0}^\infty\dfrac{x^{2r+1}}{2r+1}$

$$ \sum_{n=0}^{\infty} \frac{ (-1)^n }{(2n+1) 3^n }=\dfrac1{i/\sqrt3}\sum_{n=0}^{\infty}\dfrac{(i/\sqrt3)^{2n+1}}{2n+1}$$

Now $$2\sum_{n=0}^{\infty}\dfrac{(i/\sqrt3)^{2n+1}}{2n+1}=\log\dfrac{1+i/\sqrt3}{1-i/\sqrt3}=\log\dfrac{\sqrt3+i}{\sqrt3-i}=\log\dfrac{e^{i\pi/6}}{e^{-i\pi/6}}=\log(e^{i\pi/3})=\dfrac{i\pi}3$$

Considering principal values.

Answer 2

It is correct. From $$ \sum _{n=1}^{\infty } \frac{(-1)^n x^{2 n+1}}{2 n+1}=-x+\arctan x, \quad |x|<1, $$ you get, with $x=\dfrac1{\sqrt{3}}$: $$ \sum _{n=1}^{\infty } \frac{(-1)^n }{(2n+1)3^n}=\sqrt{3}\left(-\dfrac1{\sqrt{3}}+\arctan \dfrac1{\sqrt{3}}\right)=-1+\frac{\sqrt{3}\pi}6, $$ using $$ \arctan \dfrac1{\sqrt{3}}=\frac{\pi}6. $$

Answer 3

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#f00}{\sum_{n = 1}^{\infty}{\pars{-1}^{n} \over \pars{2n + 1}3^{n}}} & = \sum_{n = 1}^{\infty}\pars{-\,{1 \over 3}}^{n} \int_{0}^{1}x^{2n}\,\dd x = \int_{0}^{1}\sum_{n = 1}^{\infty}\pars{-\,{x^{2} \over 3}}^{n}\,\dd x = \int_{0}^{1}{-x^{2}/3 \over 1 - \pars{-x^{2}/3}}\,\dd x \\[5mm] & = -\pars{\int_{0}^{1}\,\dd x - \root{3}\int_{0}^{\root{3}/3}{\dd x \over x^{2} + 1}} = - 1 + \root{3}\arctan\pars{\root{3} \over 3} \\[5mm] & = \color{#f00}{{\root{3} \over 6}\,\pi - 1} \approx -0.0931 \end{align}