<s>[INST] My approach to determine if the following series is convergent I've just started a course in real analysis, and I don't know if I'm doing this right. Here's the exercise:

Determine if the series $$\sum_{n=1}^{\infty}\frac{3n+\sqrt{n^3+2}}{(n^2-3n+20)^2}$$ converges or diverges.

Here's what I've done so far:

$$\sum_{n=1}^{\infty}\frac{3n+\sqrt{n^3+2}}{(n^2-3n+20)^2}\\\Downarrow\\\lim_{N\to\infty}\sum_{n=1}^{N}\frac{3n+\sqrt{n^3+2}}{(n^2-3n+20)^2}\\\Downarrow\\\underbrace{\lim_{N\to\infty}\sum_{n=1}^{N}\frac{3n}{(n^2-3n+20)^2}}_{\large\text{Part 1}}+\underbrace{\lim_{N\to\infty}\sum_{n=1}^{N}\frac{\sqrt{n^3+2}}{(n^2-3n+20)^2}}_{\large\text{Part 2}} $$
From here:
Part 1
We're going to use the limit comparison test. The term $\displaystyle\ a_n=\frac{3n}{(n^2-3n+20)^2}$ behaves like $\displaystyle\ b_n=\frac{3n}{n^4}=\frac{3}{n^3}$.
We know (using the results obtained in theory) that $$\sum_{n=1}^{\infty}\frac{3}{n^3}=3\sum_{n=1}^{\infty}\frac{1}{n^3}<\infty $$.
Furthermore:
$$\lim_{n\to\infty}\displaystyle\frac{\frac{3n}{(n^2-3n+20)^2}}{\frac{3n}{n^4}}=\lim_{n\to\infty}\displaystyle\frac{n^4}{(n^2-3n+20)^2}=1$$
Then, by the limit comparison test, the series $$\sum_{n=1}^{\infty}\frac{3n}{(n^2-3n+20)^2}$$
converges.
Part 2
We do the same. Using the limit comparison test, we compare $\displaystyle\ a_n=\frac{\sqrt{n^3+2}}{(n^2-3n+20)^2}$ with $\displaystyle\ b_n=\frac{\sqrt{n^3}}{n^4}.$ If we proceed as before, we will find that $$\lim_{n\to\infty}\frac{a_n}{b_n}=1.$$
So the series converges, and as both part converges, the series $$\sum_{n=1}^{\infty}\frac{3n+\sqrt{n^3+2}}{(n^2-3n+20)^2}$$ converges.

Is this right? Am I usin the limit comparison the right way?
Thank you.
 [/INST] Note that ${3n+\sqrt{n^3+2} \over (n^2-3n+20)^2} = {n^{3 \over 2} \over n^4} { {3 \over \sqrt{n} }+ \sqrt{1+{2 \over n^3} } \over 1-{3 \over n} + {20 \over n^2} }$.
It is easy to see that ${3 \over \sqrt{n} }+ \sqrt{1+{2 \over n^3}} \le 5$ for all $n$, and for $n \ge 9$, we have 
$1-{3 \over n} + {20 \over n^2} \ge {1 \over 3}$, so
${3n+\sqrt{n^3+2} \over (n^2-3n+20)^2} \le 15 { 1\over n^{5 \over 2}} $.
Since ${5 \over 2} > 1$ we know that the series $\sum_n { 1\over n^{5 \over 2}} $ is convergent, and so by the comparison test, so is the original series.
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