Series $\sum_{n=1}^{\infty} \frac{n^2 - 5n}{n^3 + n + 1}$

Question

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?

Answer 1

Simply use the fact that$$\lim_{n\to\infty}\frac{\dfrac{n^2-5n}{n^3+n+1}}{\dfrac1n}=\lim_{n\to\infty}\frac{1-\dfrac5n}{1+\dfrac1{n^2}+\dfrac1{n^3}}=1$$and that the harmonic series diverges.

Answer 2

Diverges because $$\lim_{n\rightarrow+\infty}\frac{\frac{n^2-5n}{n^3+n+1}}{\frac{1}{n}}=1.$$

Answer 3

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.