<s>[INST] 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
 [/INST] 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
$$\tag{3}\boxed{S=\frac {\pi}{3\sqrt{3}}-\frac 12}$$
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