TITLE: Evaluation of $\sum^{\infty}_{n=1}\left(\frac{1}{3n+1}-\frac{1}{3n+2}\right)$
QUESTION [10 upvotes]: 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

REPLY [2 votes]: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}
&\bbox[10px,#ffd]{\sum_{n = 1}^{\infty}\pars{{1 \over 3n + 1} - {1 \over 3n + 2}}} = 
\sum_{n = 1}^{\infty}{1 \over \pars{3n + 2}\pars{3n + 1}}
\\[5mm] = &\
\sum_{n = 0}^{\infty}{1 \over \pars{3n + 5}\pars{3n + 4}} =
{1 \over 9}\sum_{n = 0}^{\infty}{1 \over \pars{n + 5/3}\pars{n + 4/3}}
\\[5mm] = &\
{1 \over 9}\,{\Psi\pars{5/3} - \Psi\pars{4/3} \over 5/3 - 4/3}
=
{1 \over 3}\bracks{\Psi\pars{{2 \over 3}} + {1 \over 2/3} -
\Psi\pars{{1 \over 3}} - {1 \over 1/3}}
\\[5mm] & =
-\,{1 \over 2} +
{1 \over 3}\
\underbrace{\bracks{\Psi\pars{{2 \over 3}} -
\Psi\pars{{1 \over 3}}}}_{\ds{\pi\cot\pars{\pi\,{1 \over 3}} =
{\root{3} \over 3}\,\pi}}\quad\pars{~Euler\ Reflection\ Formula~}
\\[5mm] & =
\fbox{$\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.