Appropriate test for convergence of a series when numerator and denominator have same degree? Say we have a series like this:
$$
\sum^\infty_{n=1} \frac{16n^5+2n^2+4}{6n^5+10n+1}
$$
Somehow this confused me when I saw the same power in the numerator and denominator.
My thinking: for large enough $n$, we can assume that the series behaves like $\sum \frac{16n^5}{6n^5}=\sum \frac{8}{3}$, which is divergent. So I used the Limit Comparison Test with
$$
a_n = \frac{16n^5+2n^2+4}{6n^5+10n+1}
$$
And
$$
b_n = \frac{8}{3}
$$
To find
$$
\frac{a_n}{b_n}= \frac{16n^5+2n^2+4}{6n^5+10n+1} \times \frac{3}{8} = \frac{48n^5+6n^2+12}{48n^5+80n+8} = \frac{48+6 \frac{1}{n^3} + 12 \frac{1}{n^5}}{48+ 80 \frac{1}{n^4}+8 \frac{1}{n^5}} \to 1
$$, as $n \to \infty$.
So this means the initial series is divergent.
Is that the right approach in this case when numerator and denominator have the same degree or is there something more obvious/easier that I overlooked?

We know that if a series is convergent then the sequence in the infinite sum must converge to zero. So when the sequence does not converge to zero, the series is divergent. Here, our sequence obviously converges to $\frac{16}{6}\neq 0$. Hence, the series is divergent.
