Do the following series converge or diverge? Justify. $sum_{k=0}^{infty} frac{3k^2 + 1}{k^3 + k^2 + 5}$

$$\sum_{k=0}^{\infty} \frac{3k^2 + 1}{k^3 + k^2 + 5}$$

Can I do this using direct comparison test?

for $k \in [1, \infty), a_k = \frac{3k^2 + 1}{k^3 + k^2 + 5} \geq 0$

for $k \in [1,\infty), a_k = \frac{3k^2 + 1}{k^3 + k^2 + 5} \geq \frac{3k^2}{k^3 + k^3 + 5k^3} = \frac{3}{7k} = b_k$

Consider $\frac{3}{7}\sum_{n=1}^{\infty} \frac{1}{k}$. This is a p-series with $p = 1$. By the p-series test $\sum b_k$ diverges, therefore by the comparison test $\sum a_k$ diverges too.

My textbook does this using limit comparison test wondering if I can do it using direct comparison test too. Is this right?

Answer

It's much quicker with equivalents:

$$\frac{3k^2+1}{k^3+k^2+5}\sim_\infty\frac{3k^2}{k^3}=\frac3k,\quad\text{which diverges}.$$