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

$\bf{My\; Try::}$  Here I have solved it using Definite Integration,
Like $$\sum^{\infty}_{n=1}\left(\frac{1}{3n+1}-\frac{1}{3n+2}\right)=\sum^{\infty}_{n=1}\int_{0}^{1}\left(x^{3n}-x^{3n+1}\right)dx$$
So we get $$ = \int_{0}^{1}(1-x)\sum^{\infty}_{n=1}\left(x^{3n}\right)dx = \int_{0}^{1}\frac{(1-x)x^3}{1-x^3}dx=\int_{0}^{1}\frac{x^3}{x^2+x+1}dx$$
So we get Sum $$ = \sum^{\infty}_{n=1}\left(\frac{1}{3n+1}-\frac{1}{3n+2}\right) = \int_{0}^{1}\frac{x^3}{x^2+x+1}dx = \frac{1}{18}\left(2\pi\sqrt{3}-9\right)$$
My Question is can we solve above sum without Using DEfinite Integration, If yes
Then how can I solve it, Help required
Thanks
 A: 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.
A: 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
$$\tag{3}\boxed{S=\frac {\pi}{3\sqrt{3}}-\frac 12}$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
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\begin{align}
&\bbox[10px,#ffd]{\sum_{n = 1}^{\infty}\pars{{1 \over 3n + 1} - {1 \over 3n + 2}}} = 
\sum_{n = 1}^{\infty}{1 \over \pars{3n + 2}\pars{3n + 1}}
\\[5mm] = &\
\sum_{n = 0}^{\infty}{1 \over \pars{3n + 5}\pars{3n + 4}} =
{1 \over 9}\sum_{n = 0}^{\infty}{1 \over \pars{n + 5/3}\pars{n + 4/3}}
\\[5mm] = &\
{1 \over 9}\,{\Psi\pars{5/3} - \Psi\pars{4/3} \over 5/3 - 4/3}
=
{1 \over 3}\bracks{\Psi\pars{{2 \over 3}} + {1 \over 2/3} -
\Psi\pars{{1 \over 3}} - {1 \over 1/3}}
\\[5mm] & =
-\,{1 \over 2} +
{1 \over 3}\
\underbrace{\bracks{\Psi\pars{{2 \over 3}} -
\Psi\pars{{1 \over 3}}}}_{\ds{\pi\cot\pars{\pi\,{1 \over 3}} =
{\root{3} \over 3}\,\pi}}\quad\pars{~Euler\ Reflection\ Formula~}
\\[5mm] & =
\fbox{$\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.
