<s>[INST] 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?
 [/INST] 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 $-\frac{199}{100}+\frac{\pi }{6 \sqrt{3}}+\frac{3 \log (3)}{2}\approx -0.039782$.
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