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The series is $\sum_{n=1}^{\infty}\frac{n^2+1}{n^5-n^4+3n}$.

In my book I just see examples and exercises for determining whether the series absolutely converge, conditional converge or diverge in alternating series.

This series is not alternating. So I want to make sure my analysis is right.

I have these rules:

The series $\sum a_n$ is:

  1. Absolutely convergent if $\sum |a_n|$ converges.
  2. Conditionally convergent if $\sum |a_n|$ diverges but $\sum a_n$ converges.
  3. Divergent if $\sum |a_n|$ diverges but $\sum a_n$ also diverges.

Let me know if the resume of these rules is OK.

Following these rules then:

$\sum_{n=1}^{\infty}|\frac{n^2+1}{n^5-n^4+3n}|=\sum_{n=1}^{\infty}\frac{n^2+1}{n^5-n^4+3n}$

I'm using the Direct Comparison Test:

$\frac{n^2+1}{n^5-n^4+3n}<\frac{n^2}{n^5}=\frac{1}{n^3}$

Then:

$\sum_{n=1}^{\infty}\frac{1}{n^3}$ converges because $p=3>1$ by $p$-series.

Therefore by the Direct Comparison test $\sum \frac{n^2+1}{n^5-n^4+3n}$ also converges.

Therefore $\sum \frac{n^2+1}{n^5-n^4+3n}$ is absolutely convergent.

Am I right?

Thanks in advance for your time.

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    $\begingroup$ The inequality $\frac{n^2+1}{n^5-n^4+3n}<\frac{n^2}{n^5}$ is incorrect (try specific values of $n$, or look at the numerator and denominator separately). Do you know the Limit Comparison Test? $\endgroup$ Jul 14, 2021 at 0:03
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    $\begingroup$ Basically you’re right, but just be careful on some of the details like Greg pointed out. $\endgroup$ Jul 14, 2021 at 0:09
  • $\begingroup$ @GregMartin I do know the Limit Comparison Test. But I use it when none of the criteria of the Direct Comparison test are met. I saw that $\frac{n^2+1}{n^5-n^4+3n}>\frac{n^2}{n^5}$ for n>1.175. So because my series goes from $1$ to $\infty$ and $\frac{n^2+1}{n^5-n^4+3n}\nless\frac{n^2}{n^5}$ in the whole interval $[1,\infty)$ I must use the Limit Comparison Test? $\endgroup$
    – bdvg2302
    Jul 14, 2021 at 0:23
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    $\begingroup$ I wouldn't think of it as "I must use the limit comparison test", but rather "my current argument is mathematically incorrect, so I need to change it". The LCT is my suggestion for a tool that aligns most closely with your current thinking but doesn't fall prey to the same flaw. $\endgroup$ Jul 14, 2021 at 0:26
  • $\begingroup$ @GregMartin I'm sorry but I'm not following. Change my argument exactly to what? If I use the LCT it would be $\lim_{x \to \infty} \frac{\frac{n^2+1}{n^5-n^4+3n}}{\frac{1}{n^3}}$. The result of it would be 1 and because is finite and positive and because $\sum_{n=1}^{\infty}\frac{1}{n^3}$ converges by the p-series Test, then both converges. Therefore, $\sum_{n=1}^{\infty}\frac{n^2+1}{n^5-n^4+3n}$ is absolutely convergent? $\endgroup$
    – bdvg2302
    Jul 14, 2021 at 1:40

1 Answer 1

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First off, your series isn't alternating, so you cannot use the Alternating Series Test. The terms "absolutely\conditionally converge" ONLY apply to alternating series. If your series is not alternating, it either just converges or diverges. These terms don't apply. You could say it's absolutely convergent, because the absolute value doesn't affect the value of the sum, but this is rather unnecessary.

Notice you have a polynomial of degree $2$ on the numerator, and a polynomial of degree $5$ in the denominator. So, as $n$ gets very large, this fraction looks more and more like $\frac{n^2}{n^5} = \frac{1}{n^3}$. Here is a graph with $\frac{1}{n^3}$ in red and $\frac{n^2+1}{n^5-n^4+n}$ in blue. Notice the functions converge to zero together as $n$ gets large. This is the idea behind the Limit Comparison Test, which goes like this:

Suppose we have two series $\sum a_n$ and $\sum b_n$, and let $L = \lim_{n\to\infty} \frac{a_n}{b_n}$. If $L$ is finite and positive, then $a_n$ and $b_n$ either BOTH converge or BOTH diverge. Otherwise, we cannot come to any conclusion.

So, take $a_n$ = $\frac{1}{n^3}$ and $b_n = \frac{n^2+1}{n^5-n^4+n}.$ Then,

$$\frac{a_n}{b_n} = \frac{\frac{1}{n^3}}{\frac{n^2+1}{n^5-n^4+n}} = \frac{n^5 - n^4 + 1}{n^5+n^3}$$

And, when we take the limit, we see that

$$L = \lim_{n\to\infty} \frac{a_n}{b_n} = \lim_{n\to\infty} \frac{n^5 - n^4 + 1}{n^5+n^3} = 1$$

Because $L=1$, it is positive and finite, and we conclude that BOTH $\sum\frac{1}{n^3}$ and $\sum\frac{n^2+1}{n^5-n^4+n}$ converge, or both diverge. Because $\sum\frac{1}{n^3}$ converges, we know that our given sum, $\sum\frac{n^2+1}{n^5-n^4+n}$, must also converge.

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