Does $\sum^\infty_{k=1} \frac{k^3-2k-3}{k^4}$ converge?

Question

Does $\sum^\infty_{k=1} \frac{k^3-2k-3}{k^4}$ converge?

I tried with the Integral test for convergence, and I got

$ln|x|+ \frac{k+1}{k^3}$ from 1 to $\infty$

Does this integral exist and how to show that?

Answer 1

HINT

We have

$$\sum^\infty_{k=1} \frac{k^3-2k-3}{k^4} =\sum^\infty_{k=1} \frac1k-2\sum^\infty_{k=1} \frac{1}{k^3}-3\sum^\infty_{k=1} \frac{1}{k^4}$$

then recall that the last two converge.

Answer 2

By the fundamental theorem of algebra there exist roots $r_1$, $r_2$ and $r_3$ such that $$ k^3-2k-3=(k-r_1)(k-r_2)(k-r_3). $$ And so we can write $$ \frac{k^3-2k-3}{k^4} = \frac{1}{k}\frac{(k-r_1)}{k}\frac{(k-r_2)}{k}\frac{(k-r_3)}{k} = \frac{1}{k}\left(1-\frac{r_1}{k} \right)\left(1-\frac{r_2}{k} \right)\left(1-\frac{r_3}{k} \right) $$ It's easy to prove (this is an easy exercise I leave to you) that $\sum_{k=1}^{\infty}a_k=\infty$ and $\lim_{k\to \infty}b_k=1$ implies $\sum_{k=1}^{\infty}a_kb_k=\infty$. So, just put $a_k=\frac{1}{k}$ and $b_k=\left(1-\frac{r_1}{k} \right)\left(1-\frac{r_2}{k} \right)\left(1-\frac{r_3}{k} \right)$ and get $$ \sum^\infty_{k=1} \frac{k^3-2k-3}{k^4} = \lim_{k\to\infty}\sum_{k=1}^{\infty}\frac{1}{k}\left(1-\frac{r_1}{k} \right)\left(1-\frac{r_2}{k} \right)\left(1-\frac{r_3}{k} \right)=\infty $$