Compute the following sum :
S=$\sum_{n=1}^{\infty}\frac{1}{6n^5+15n^4+10n^3-n}$
My attempt : Use partial fraction :
$6n^5+15n^4+10n^3-n=n(n+1)(2n+1)(3n^2+3n-1)$ $S=\sum_{n=1}^{\infty}(\frac{9(2n+1)}{7(3n^2+3n-1)}-\frac{1}{n}-\frac{1}{1+n}+\frac{16}{7(2n+1)})$ Then use identity digamma " sum " But I find divergence sum
Hint $$6n^5+15n^4+10n^3-n=6n(n+1)\left(n+\frac12 \right)\left(n+\frac{\sqrt{21}+3}{6} \right)\left(n-\frac{\sqrt{21}-3}{6} \right)$$ Use partial fraction decomposition and generalized harmonic numbers or polygamma functions.
Edit
Whatever the denominator will be, after partial fraction decomposition, you have $$\frac 1 {P_k(n)}=\sum_{i=1}^k \frac {a_i}{n-r_i}$$ and $$S_p=\sum_{n=1}^p \frac {1}{n-r_i}=\psi (p+1-r_i)-\psi (1-r_i)$$ Now, using asymptotics $$S_p=\log (p)-\psi (1-r_i)+\frac{1-2r_i}{2p}-\frac{6 r_i^2-6 r_i+1}{12 p^2}+O\left(\frac{1}{p^3}\right)$$ You just need to recombine everything and continue with Taylor expansions to get not only the limit but also how it is approached.
For sure, the limit will be finite only if $\sum_{i=1}^k a_i=0$.
As an example, consider $$\sum_{n=1}^\infty \frac{1}{\left(n+\frac{2}{3}\right) \left(n+\frac{3}{4}\right) \left(n+\frac{5}{6}\right)}$$ $$\frac{1}{\left(n+\frac{2}{3}\right) \left(n+\frac{3}{4}\right) \left(n+\frac{5}{6}\right)}=\frac{216}{3 n+2}-\frac{576}{4 n+3}+\frac{432}{6 n+5}$$ $$\sum_{n=1}^p\frac{1}{\left(n+\frac{2}{3}\right) \left(n+\frac{3}{4}\right) \left(n+\frac{5}{6}\right)}=72 \psi \left(p+\frac{5}{3}\right)-144 \psi \left(p+\frac{7}{4}\right)+72 \psi \left(p+\frac{11}{6}\right)-72 \psi \left(\frac{11}{6}\right)+144 \psi \left(\frac{7}{4}\right)-72 \psi \left(\frac{5}{3}\right)$$ So, using the asymptotics, for infintely large values of $p$ $$\sum_{n=1}^p\frac{1}{\left(n+\frac{2}{3}\right) \left(n+\frac{3}{4}\right) \left(n+\frac{5}{6}\right)}=-72 \left(\psi \left(\frac{5}{3}\right)-2 \psi \left(\frac{7}{4}\right)+\psi \left(\frac{11}{6}\right)\right)-\frac{1}{2 p^2}+O\left(\frac{1}{p^3}\right)$$