Prove $\sum_{k=0}^\infty \frac{(-1)^k}{3k+2} = \frac{1}{9}\left(\sqrt{3}\pi-3\ln\,2\right)$

Question

Just for fun:

How can we prove (calculate) that $\sum_{k=0}^\infty \frac{(-1)^k}{3k+2} = \frac{1}{9}\left(\sqrt{3}\pi-3\ln\,2\right)$ ? Can we use (9) from:

http://mathworld.wolfram.com/DigammaFunction.html

(9): $\sum_{k=0}^\infty \frac{(-1)^k}{3k+1} = \frac{1}{9}\left(\sqrt{3}\pi+3\ln\,2\right)$

?

Thx!

Answer 1

We can evaluate the series of interest without appealing to the Digamma Function.

Note that we can simply write

$$\begin{align} \sum_{n=0}^{2N}\frac{(-1)^{n}}{3n+2}&=\sum_{n=0}^N\left(\frac{1}{6n+2}-\frac{1}{6n+5}\right)\\\\ &=\sum_{n=0}^N\int_0^1\left( x^{6n+1}-x^{6n+6}\right)\,dx\\\\ &=\int_0^1 x\left(\frac{1-x^{6N+6}}{1+x^3}\right)\,dx\\\\ &=\int_0^1 \left(\frac{x}{1+x^3}\right)\,dx -\int_0^1 x^{6N+7} \left(\frac{1}{1+x^3}\right)\,dx \end{align}$$

Applying the Dominated Convergence Theorem (or alternatively, integrate by parts and observe that the second integral is $ O(N^{-1})$), we find that

$$\sum_{n=0}^\infty \frac{(-1)^{n}}{3n+2}=\int_0^1 \frac{x}{1+x^3}\,dx$$

Can you finish now using partial fraction expansion for example?

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{k = 0}^{\infty}{\pars{-1}^{k} \over 3k + 2} & = \sum_{k = 0}^{\infty}{\pars{\expo{\ic\pi/3}}^{3k} \over 3k + 2} = \sum_{k = 0}^{\infty}{\pars{\expo{\ic\pi/3}}^{k} \over k + 2} \,{1 + \expo{2k\pi\ic/3} + \expo{-2k\pi\ic/3} \over 3} \\[5mm] & = {1 \over 3}\sum_{k = 0}^{\infty}{\expo{\ic k\pi/3} + \pars{-1}^{k} + \expo{-\ic k\pi/3} \over k + 2} = {2 \over 3}\,\Re\sum_{k = 0}^{\infty}{\expo{\ic k\pi/3} \over k + 2} + {1 \over 3}\,\Re\sum_{k = 0}^{\infty}{\pars{-1}^{k} \over k + 2} \\[5mm] & = {2 \over 3}\,\Re\pars{\expo{-2\pi\ic/3}\sum_{k = 2}^{\infty}{\expo{\ic k\pi/3} \over k}} + {1 \over 3}\,\Re\sum_{k = 2}^{\infty}{\pars{-1}^{k} \over k} \\[5mm] & = {2 \over 3}\,\Re\pars{-\expo{-\ic\pi/3} + \expo{-2\pi\ic/3}\sum_{k = 1}^{\infty}{\expo{\ic k\pi/3} \over k}} + {1 \over 3}\pars{1 + \sum_{k = 1}^{\infty}{\pars{-1}^{k} \over k}} \\[5mm] & = {2 \over 3}\braces{-\,{1 \over 2} - \Re\bracks{\expo{-2\pi\ic/3}\ln\pars{1 - \expo{\ic\pi/3}}}} + {1 \over 3}\braces{1 - \ln\pars{1 - \bracks{-1}}} \\[5mm] & = -\,{1 \over 3}\ln\pars{2} - {2 \over 3}\,\Re\bracks{% -\,{1 + \root{3}\ic \over 2} \ln\pars{{1 \over 2} - {\root{3} \over 2}\,\ic}} \\[5mm] & = -\,{1 \over 3}\ln\pars{2} - {2 \over 3}\,\Re\bracks{% -\,{1 + \root{3}\ic \over 2} \pars{-\,{\pi \over 3}\,\ic}} \\[5mm] & = \bbx{-\,{1 \over 3}\ln\pars{2} + {\root{3} \over 9}\,\pi} \approx 0.3736 \end{align}