Evaluation of $\sum^{\infty}_{n=1}\left(\frac{1}{3n+1}-\frac{1}{3n+2}\right)$

An Eulerian approach. The function $\sin(\pi x)$ has its zeroes at the integers, hence the function $f(x)=\sin\left(\frac{\pi}{3}(x+2)\right)$ has its zeroes at $\{ \ldots,-8, -5,-2,1,4,7,\ldots \}$. Since the Taylor series of $f(x)$ at $x=0$ is given by: $$ f(x) = \frac{\sqrt{3}}{2}-\frac{\pi}{6}x+O(x^2) $$ it happens that:

$$ \lim_{n\to +\infty}\sum_{k=-n}^{n}\frac{1}{3k+1} = -\frac{[x^1]\,f(x)}{[x^0]\,f(x)} = \color{red}{\frac{\pi}{3\sqrt{3}}} $$

and the claim easily follows. We may regard $f(x)$ as "an infinite-degree-polynomial" since $\sin(z)$ is an entire function whose Weierstrass product has no exponential term. With the same approach you may prove the more general identity: $$ \sum_{n\geq 0}\left(\frac{1}{kn+1}-\frac{1}{k(n+1)-1}\right)=\frac{\pi}{k}\,\cot\left(\frac{\pi}{k}\right)$$ that also follows from the reflection formula for the digamma function.

Well you could rewrite your sum as : $$\tag{1}S=-\frac 12+\frac 1{\sin(2\pi/3)}\;\sum_{k=1}^\infty\frac {\sin(2\pi k/3)}{k}$$ (the $-\dfrac 12$ is from your sum starting at $n=1$ and not $n=0$)
and use the Fourier series for the sawtooth wave to get your result

or consider $(1)$ as $-\frac 12$ plus the imaginary part of : $$\frac 1{\sin(2\pi/3)}\;\sum_{k=1}^\infty\frac {\exp(2\pi k\,i/3)}{k}=\frac 1{\sin(2\pi/3)}\;\sum_{k=1}^\infty\frac {\left(\exp(2\pi \,i/3)\right)^{\;k}}{k}$$ that is \begin{align} \tag{2}S&=-\frac 12-\frac 2{\sqrt{3}}\;\Im\;\log(1-\exp(2\pi \,i/3))\\ &=-\frac 12-\frac 1{\sqrt{3}\;i}\;\log\frac{1-\exp(2\pi \,i/3)}{1-\exp(-2\pi \,i/3)}\\ &=-\frac 12-\frac 1{\sqrt{3}\;i}\;\log(\exp(-\pi \,i/3))\\ \end{align} and thus $$\tag{3}\boxed{S=\frac {\pi}{3\sqrt{3}}-\frac 12}$$

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\sum_{n = 1}^{\infty}\pars{{1 \over 3n + 1} - {1 \over 3n + 2}}} = \sum_{n = 1}^{\infty}{1 \over \pars{3n + 2}\pars{3n + 1}} \\[5mm] = &\ \sum_{n = 0}^{\infty}{1 \over \pars{3n + 5}\pars{3n + 4}} = {1 \over 9}\sum_{n = 0}^{\infty}{1 \over \pars{n + 5/3}\pars{n + 4/3}} \\[5mm] = &\ {1 \over 9}\,{\Psi\pars{5/3} - \Psi\pars{4/3} \over 5/3 - 4/3} = {1 \over 3}\bracks{\Psi\pars{{2 \over 3}} + {1 \over 2/3} - \Psi\pars{{1 \over 3}} - {1 \over 1/3}} \\[5mm] & = -\,{1 \over 2} + {1 \over 3}\ \underbrace{\bracks{\Psi\pars{{2 \over 3}} - \Psi\pars{{1 \over 3}}}}_{\ds{\pi\cot\pars{\pi\,{1 \over 3}} = {\root{3} \over 3}\,\pi}}\quad\pars{~Euler\ Reflection\ Formula~} \\[5mm] & = \fbox{$\ds{{\root{3} \over 9}\,\pi - \half}$} \approx 0.1046 \end{align}

$\Psi\pars{z}$ is the digamma function where we used its recurrence relation and the Euler identity.