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.

Update: This uses integration, which is something OP sought to avoid.
$$\begin{split}
\sum_{n=0}^\infty \left( \frac{1}{3n+4}-\frac{3}{3n+2}+\frac{2}{3n+1}\right)
&=\sum_{n=0}^\infty \int_0^1x^{3n}(x^3-3x+2)dx\\
&= \int_0^1\frac{x^3-3x+2}{1-x^3}dx\\
&= \int_0^1 \left( \frac 3 {x+x^2+1}-1\right)dx\\
&= \left[ 2{\sqrt 3}\arctan\left(\frac{2x+1}{\sqrt 3}\right) - x\right]_0^1 \\
&= 2{\sqrt 3} \left( \frac \pi 3 - \frac \pi 6\right)-1\\
&= \frac{\pi}{\sqrt 3}-1
\end{split}$$
