question about applying a comparison test to a sequence In lecture, we were asked:
Does $ \sum \limits_{n=0}^\infty \frac{ 2n^3+ 3n -8 }{ n^5-5n^3-n^2+2  }$ converge?
$a_n = \frac{ 2n^3+ 3n -8 }{ n^5-5n^3-n^2+2  }$
In discussing strategies for applying a comparison test for $a_n$ we rejected several options for choosing a bound for $b_n$. The strategy is to choose an expression larger in the numerator and smaller in the denominator for $b_n$  If I understand correctly, using $2n^5$ in the denominator for $b_n$ doesn't work, because, as n gets large, we have
$$ \frac{2n^3}{n^5} \not < \frac{3n^3}{2n^5}$$
$$ 2 \not < \frac{3}{2} $$
If this is the case, couldn't we choose $b_n = \frac{10n^3}{2n^5}$?
Plugging the first inequality into Wolfram Alpha was not enlightening.


From the lecture:
To apply the comparison test, we need to bound the numerator from above and the denominator from below.
For the numerator of $b_n$, $2n^3$ won't work because we have $3n$ in the numerator of $a_n$. $3n^3$ is large enough.
$2n^5$ in the denominator of $b_n$ won't work. 
$a_n = \frac{ 2n^3+ 3n -8 }{ n^5-5n^3-n^2+2  } \ \substack{ <\\ \\ >} \ \frac{3n^3}{2n^5} =b_n$
Using $n^4$ in the denominator won't work because this gives us a harmonic series for $b_n$, which is divergent.
$a_n = \frac{ 2n^3+ 3n -8 }{ n^5-5n^3-n^2+2  } \ \substack{ <\\ \\ >} \ \frac{3n^3}{n^4}=b_n $
Using asymptotic analysis, we get $a(x)= \frac{2}{x^2} + \mathcal{O}(\frac{1}{x^4})$, which gives us an integral with p=2, which converges. 

We do an explicit comparison, since that's what the question seems to ask for, but there are simpler ways.
For $n\gt 1$, the top is positive and  $\lt 2n^3+3n^3=5n^3$.
The bottom is trickier. Rewrite it as
$$\frac{1}{2}n^5+ \frac{1}{2}n^5-5n^3-n^2+2.$$
Note that if $n\gt 10$, then $\frac{1}{2}n^5-5n^3-n^2+2\gt 0$. This is because then $\frac{1}{2}n^5 \gt 50n^3$. (We could get away with something cheaper than $10$, but why bother?). So the bottom, when $n\gt 10$, is $\gt \frac{1}{2}n^5$.
What we did here was to despatch half  of $n^5$ to go crush $5n^3+n^2$. After they are crushed, the bottom is bigger than the remaining $\frac{1}{2}n^5$. 
So for $n\gt 10$ we have that our expression is positive and $\lt \frac{5n^3}{x^5/2}=\frac{10}{n^2}$. Good enough to show convergence.
