Series $\sum_{n=1}^{\infty} \frac{n^2 - 5n}{n^3 + n + 1}$ Alright, this is another problem that I have been stuck on. The goal is to determine whether it is convergent or divergent.
$$\sum_{n=1}^{\infty} \frac{n^2 - 5n}{n^3 + n + 1}$$
So to start off, Integral Test seems rough as the denominator is not factorable for partial fraction decomposition.
So then, I tried Direct Comparison Theorem, but...
$$\frac{1}{n^3} < \frac{n^2 - 5n}{n^3 + n + 1}[n > 5]$$ 
Although not for the intervals [0, 5].
$$\frac{1}{n^3} > \frac{n^2 - 5n}{n^3 + n + 1}[0<n<5]$$ 
So yeah, that's kind of confusing. Especially since the problem starts at n = 1 instead of n = 5.
However, I know that, by p-series
$$\sum_{n=1}^{\infty} \frac{1}{n^3} --> converges $$ 
And if the smaller value converges, then Direct Comparison Theorem tells us nothing.
So I decided to try the Limit Comparison Theorem:
$$b_n = \frac{1}{n^3} $$
$$\lim_{n\to0} \frac{n^2 - 5n}{n^3+n+1}*\frac{n^3}{1} = \lim_{n\to0} \frac{n^6 - 5n^4}{n^3 + n + 1} = {\infty}$$
So if bn is convergent by p series, but the limit is divergent, then LCT is useless.
So, now my question is where did I go wrong in attempting to prove convergence/divergence?

For $n \ge 5$, $$\frac{n^2 - 5n}{n^3 + n + 1} > \frac{(n^2 + 2n + 1) - (7n + 7) + 6}{n^3 + 3n^2 + 3n + 1} > \frac{(n+1)^2 - 7(n+1)}{(n+1)^3} = \frac{1}{n+1} - \frac{7}{(n+1)^2}.$$  Since $$\sum_{n=5}^\infty \frac{1}{n+1}$$ does not converge but $$\sum_{n=5}^\infty \frac{7}{(n+1)^2}$$ is convergent, the original sum does not converge.
