<s>[INST] Find $\sum_{n=0}^\infty \big( \frac{1}{3n+4}-\frac{3}{3n+2}+\frac{2}{3n+1}\big)$ without power series or integration? In a calculus, I have to find a direct way to compute the sum
$$\sum_{n=0}^\infty \left( \frac{1}{3n+4}-\frac{3}{3n+2}+\frac{2}{3n+1}\right).$$
I tried to reform a partial sum:
$$\sum_{n=0}^N a_n = \frac12+3\sum_{n=1}^N \left(\frac{1}{3n+1}-\frac{1}{3n+2}\right).$$
A classic way then is to introduce the missing terms to use Euler's formula. Problem is, I don't know how to deal with the alternating signs.
What I can't do: use power series (the sum comes from one), neither integration (same problem). I need to find a direct way to sum this, that is, a way which only involves classic techniques with series.
Anyone to help me? Thanks.
 [/INST] First note that
$$\sum_{n=0}^\infty z^n = \frac{1}{1-z}$$
implies
$$\sum_{n=1}^\infty \frac{z^n}{n} = \log\frac{1}{1-z} = -\log(1-z),$$
and so
\begin{align}
\sum_{n=0}^\infty \frac{z^n}{n+1} &= \frac{-\log(1-z)}{z} \tag1\\
\sum_{n=0}^\infty \frac{z^n}{n+2} &= \frac{-\log(1-z)-z}{z^2} \tag2\\
\sum_{n=0}^\infty \frac{z^n}{n+4} &= \frac{-6\log(1-z)-6z-3z^2-2z^3}{6z^4} \tag3
\end{align}
Now let
$$a_n = \frac{1}{n+4}−\frac{3}{n+2}+\frac{2}{n+1}$$
and apply $(1)$ through $(3)$ to yield
$$\sum_{n=0}^\infty a_n z^n = \sum_{n=0}^\infty \frac{z^n}{n+4} − 3\sum_{n=0}^\infty \frac{z^n}{n+2} + 2\sum_{n=0}^\infty \frac{z^n}{n+1}
= \frac{(-6+18z^2-12z^3)\log(1-z) - 6z - 3z^2 + 16z^3}{6 z^4}.$$
Finally apply
$$\sum_{n=0}^\infty a_{3n} = \sum_{n=0}^\infty a_n\frac{1+\omega^n+\omega^{2n}}{3} =  \frac{1}{3}\sum_{n=0}^\infty a_n + \frac{1}{3}\sum_{n=0}^\infty a_n\omega^n + \frac{1}{3}\sum_{n=0}^\infty a_n(\omega^2)^n,$$
where $\omega=\exp(2\pi i/3)$,
to obtain
$$\frac{1}{3}\cdot\frac{7}{6} + \frac{1}{3}\left(\frac{-25-19 i\sqrt3}{12} + 3 i \sqrt3 \log\left(\frac{3 - i \sqrt3}{2}\right)\right) + \frac{1}{3}\left(\frac{-25 + 19 i\sqrt3}{12} - 3 i \sqrt3 \log\left(\frac{3 + i \sqrt3}{2}\right)\right) = \frac{\pi}{\sqrt3} - 1.$$
</s>