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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.

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    $\begingroup$ It is done the right way. There was no particular need to separate the sum into two parts. Limit Comparison with $\sum_1^\infty \frac{n^{3/2}}{n^4}$ will do it in one step. $\endgroup$ – André Nicolas Sep 28 '15 at 17:05
  • $\begingroup$ @AndréNicolas In my notes, for using the limit comparison test, I have as a condition that $a_n \leq Cb_n$, being $C$ a fixed number. How do I know that $a_n$ is bounded by $b_n$ to be sure that I'm using the "right" $b_n$? $\endgroup$ – Relure Sep 28 '15 at 17:08
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    $\begingroup$ The terms are positive. If you can show that the limit of the ratio is a positive constant, you are finished. In this case the limit of the ratio is $1$. $\endgroup$ – André Nicolas Sep 28 '15 at 17:17
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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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