Calculating sum of converging series $\sum_{n=1}^{\infty}\frac{1-n}{9n^3-n}$ I have a trouble with calculating the sum of this series:
$$2+\sum_{n=1}^{\infty}\frac{1-n}{9n^3-n}$$
I tried to split it into three separate series like this: 
$$2+\sum_{n=1}^{\infty}\frac{1-n}{9n^3-n} =2+\sum_{n=1}^{\infty}\frac{2}{3n+1}+\sum_{n=1}^{\infty}\frac{1}{3n-1}-\sum_{n=1}^{\infty}\frac{1}{n} $$ but I'm not able to continue, can you give me some tips?

For the direct evaluation of the limit, you have received the good solution from J.G.
You could also consider the partial sum using, as you did, partial fraction decomposition
$$\frac{1-n}{9n^3-n}=\frac{1}{3 n-1}+\frac{2}{3 n+1}-\frac{1}{n}$$ which makes
$$S_p=\sum_{n=1}^{p}\frac{1-n}{9n^3-n}=\frac{1}{3} \left(\psi \left(p+\frac{2}{3}\right)-\psi
   \left(\frac{2}{3}\right)\right)+\frac{2}{3} \left(\psi
   \left(p+\frac{4}{3}\right)-\psi \left(\frac{4}{3}\right)\right)-H_p$$ where appear the digamma function and harmonic number.
Now, using asymptotics and Taylor series you should arrive to
$$S_p=-\left(\gamma +\frac{1}{3}\psi\left(\frac{2}{3}\right)+\frac{2}{3} \psi
  \left(\frac{4}{3}\right)\right)+\frac{1}{9 p}-\frac{1}{9
   p^2}+O\left(\frac{1}{p^3}\right)$$ which shows the limit and how it is approached.
Now, the particular values
$$\psi\left(\frac{2}{3}\right)=-\gamma +\frac{\pi }{2 \sqrt{3}}-\frac{3 \log (3)}{2}$$
$$\psi\left(\frac{4}{3}\right)=3-\gamma -\frac{\pi }{2 \sqrt{3}}-\frac{3 \log (3)}{2}$$ make
$$S_p=\left(-2+\frac{\pi }{6 \sqrt{3}}+\frac{3 \log (3)}{2}\right)+\frac{1}{9
   p}-\frac{1}{9 p^2}+O\left(\frac{1}{p^3}\right)$$
For $p=10$, the exact value is $-\frac{477820712081}{12033629407800}\approx -0.039707$ while the above truncated expansion gives $-\frac{199}{100}+\frac{\pi }{6 \sqrt{3}}+\frac{3 \log (3)}{2}\approx -0.039782$.
