convergence $\sum^{\infty}_{k=1}\frac{k^2+3k+1}{k^3-2k-1}$ 
Finding whether the series $$\sum^{\infty}_{k=1}\frac{k^2+3k+1}{k^3-2k-1}$$  is converges or diverges.

What i try
$$\frac{k^2+3k+1}{k^3-2k-1}\approx\frac{k^2}{k^3}=\frac{1}{k}$$
So our series seems to ne diverges.
But i did not understand How do i use inequality sign here. So that i can justify my answe. Help me please. Thanks
 A: $\frac {k^{2}+3k+1} {k^{3}-2k-1} >\frac {k^{2}} {k^{3}}= \frac  1 k \:\forall k\in\mathbb{N} $ and $\sum \frac 1k $ is divergent. Therefore $\sum_{k=1}^{\infty}\frac {k^{2}+3k+1} {k^{3}-2k-1}$ diverges by the comparison test.
[Note that $k^{2}+3k+1 >k^{2}$ and $k^{3}-2k-1 <k^{3} \:\forall k\in\mathbb{N}$. This justifies the inequality above]. 
A: Other than using comparison test, you may use the limit comparison test as follows: 
$$\lim_{k\to\infty}\frac{\left(\frac{k^2+3k+1}{k^3-2k-1}\right)}{\left(\frac{1}{k}\right)}=\lim_{k\to\infty}\frac{k^3+3k^2+k}{k^3-2k-1}
=\lim_{k\to\infty}\frac{1+\frac{3}{k}+\frac{1}{k^2}}{1-\frac{2}{k^2}-\frac{1}{k^3}}=1\neq 0.$$
Since the harmonic series $\displaystyle\sum_{k=1}^\infty\frac{1}{k}$ diverges, 
by limit comparison test, 
$\displaystyle\sum^{\infty}_{k=1}\frac{k^2+3k+1}{k^3-2k-1}$ also diverges. 
A: It suffices to show that for sufficiently large $k$,
$$\frac{k^2+3k+1}{k^3-2k-1} > \frac{a}{k}$$
for some positive real $a$. If we set $a = \frac 12$, then our inequality becomes
$$2k^3 + 6k^2 + 2k > k^3-2k-1,$$
which upon rearranging gives
$$k^3 + 6k^2 + 4k + 1 > 0,$$
as long as $k^3-2k-1>0$ (which is true for all $k \geq 2$). This inequality clearly true for all positive $k$. So we have
$$\frac{k^2+3k+1}{k^3-2k-1} > \frac{1}{2k}$$
for any $k \geq 2$,
and then you can use the comparison test on the harmonic series.
