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I have the next sequence queue: $\sum_{n=1}^{\infty}{\frac{n^2-n-1}{n^4+n^2+1}}$. Does the queue converges or diverge?


My attempt: I have tried to show that $\frac{1}{n^2}>|\frac{n^2-n-1}{n^4+n^2+1}|$ for any $n>0$, and then by the comparation test we get that since $\sum_{n=1}^{\infty}{\frac{1}{n^2}}$ converges, we have that $\sum_{n=1}^{\infty}{|\frac{n^2-n-1}{n^4+n^2+1}|}$ converges too, and by a theorem we have that since $\sum_{n=1}^{\infty}{|\frac{n^2-n-1}{n^4+n^2+1}|}$ $\implies$ $\sum_{n=1}^{\infty}{\frac{n^2-n-1}{n^4+n^2+1}}$ converges. Is that right?

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  • $\begingroup$ Yeah, a direct comparison test works for this. You were right to compare the following: $$\frac{1}{n^2}>\left|\frac{n^2-n-1}{n^4+n^2+1}\right|$$ Now because the $$\sum_{n=1}^{\infty} \frac{1}{n^2}$$ converges by the p-test, you can conclude that the original series also converges. $\endgroup$ Mar 16, 2021 at 22:22
  • $\begingroup$ @ObsessiveInteger So my attempt is right? but can you show me how I satisfy that $\frac{1}{n^2}>|\frac{n^2-n-1}{n^4+n^2+1}|$. for $\frac{1}{n^2}>\frac{n^2-n-1}{n^4+n^2+1}$ it's easy since we get by algebra that $n^3+1>0$ for $n>0$. However, for the negative it's not that easy. $\endgroup$
    – Chopin
    Mar 16, 2021 at 22:31
  • $\begingroup$ You could make an argument by saying that the numerator of $\frac{n^2-n-1}{n^4+n^2+1}$ grows smaller faster compared to that of $\frac{1}{n^2}$ and similarly, the denominator of $\frac{n^2-n-1}{n^4+n^2+1}$ grows bigger much faster compared to that of $\frac{1}{n^2}$. This means that $\frac{n^2-n-1}{n^4+n^2+1}$ is reaching $0$ faster compared to $\frac{1}{n^2}$, hence, the inequality is true. Of course you can see that graphically as well. Plug in both functions into desmos and you will see. $\endgroup$ Mar 16, 2021 at 22:41
  • $\begingroup$ @ObsessiveInteger seeing that throw desmos is good, but can I actually verify that by using just inequality algebra? and thank you for the answer. $\endgroup$
    – Chopin
    Mar 16, 2021 at 22:44
  • $\begingroup$ Why don't you do the limit comparison test instead? That way you do not have to worry about proving the inequality. $\endgroup$ Mar 16, 2021 at 22:45

2 Answers 2

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I continue from where you stopped,

Easy to show that :

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

When we know that $\sum_{n=1}^{\infty}{\frac{1}{n^2}}$ converges.

Therefore, by comparison test :

$$\underset{converges}{\underbrace{\frac{n^2-n-1}{n^4+n^2+1}}} < \underset{converges}{\underbrace{\frac{1}{n^2}}}$$

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  • $\begingroup$ However, what stops us from doing so for just the expression without ||, is that when $n$ has small values then the whole expression is negative, and then you cannot use the comparison test, so we have to show it with ||. $\endgroup$
    – Chopin
    Mar 16, 2021 at 22:48
  • $\begingroup$ @aasc232 By the limit test we get the equation aims to zero so can see that only the first element in the sum has negative value do not forget is the sum of this equation . $\endgroup$
    – ATB
    Mar 16, 2021 at 22:54
  • $\begingroup$ Can you remind me the limit test? $\endgroup$
    – Chopin
    Mar 16, 2021 at 22:56
  • $\begingroup$ And the comparison test is when you have $0\le a_k\le b_k$ for all $k>0$, with the first elements of $a_k$ included, then if $b_k$ converges we get that $a_k$ converges too. So since we have that $a_k$ isn't positive for all $k>0$ we cannot use it. That is what I tried telling you. Isn't right? $\endgroup$
    – Chopin
    Mar 16, 2021 at 23:01
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    $\begingroup$ @aasc232 Do the $n=1$ term separately $$|\frac{1^2-1-1}{1^{4}+1^2+1}|<\frac{1}{1}$$ and then ATB's answer applies for all $n>1$ as $\frac{n^{2}-n-1}{n^{4}+n^{2}+1}$ is positive for $n>1$. Of course, a finite number of terms does not determine the convergence and can be disregarded in general when checking convergence $\endgroup$
    – user649348
    Mar 16, 2021 at 23:03
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We are trying to prove that the series, $$\sum_{n=1}^{\infty}{\frac{n^2-n-1}{n^4+n^2+1}}$$ converges.

Let us use the Limit Comparison Test. That is, if $a_n>0$ and $b_n>0$, then if $$\lim_{n\to \infty}\frac{a_n}{b_n}=L$$ for some real number $L$. Then, if $b_n$ converges, by the limit comparison test, $a_n$ must also converge.

Let $a_n=\frac{n^2-n-1}{n^4+n^2+1}$ and we say $b_n=\frac{1}{n^2}$. Then, $$\lim_{n\to \infty} \frac{\frac{n^2-n-1}{n^4+n^2+1}}{\frac{1}{n^2}}\\ \lim_{n\to \infty}\left(\frac{(n^2-n-1)(n^2)}{n^4+n^2+1}\right)\\ \lim_{n\to \infty}\frac{n^4-n^3-n^2}{n^4+n^2+1}\cdot \frac{\frac{1}{n^4}}{\frac{1}{n^4}}\\ \lim_{n\to \infty} \left(\frac{{1-\frac{1}{n}}-\frac{1}{n^2}}{1+\frac{1}{n^2}+\frac{1}{n^4}}\right)\\ \lim_{n\to \infty}\frac{1-0-0}{1+0+0}=\frac{1}{1}=1$$

Hence, we have shown that the $\lim_{n\to \infty}\frac{a_n}{b_n}=L$ converges because it equals some constant; in this case, $1$.

Now, because $\sum_{n=1}^{\infty}\frac{1}{n^2}$ converges by the p-test, (since $p=2>1$), we can conclude that the series, $$\sum_{n=1}^{\infty}{\frac{n^2-n-1}{n^4+n^2+1}}$$ must also converge by the limit comparison test.

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  • $\begingroup$ Ohhh I didn't try using this, in this way. Thank you! $\endgroup$
    – Chopin
    Mar 16, 2021 at 23:03
  • $\begingroup$ One last thing, by limit comparison test we have to show that the sequence queue $\frac{n^2−n−1}{n^4+n^2+1}$ is positive. How do we show that? or you saying we disregard the first negative elements of the sum?' $\endgroup$
    – Chopin
    Mar 16, 2021 at 23:12
  • $\begingroup$ Technically you do not have to show that, note that the sum $\sum_{n=1}^{\infty}$ starts at $n=1$, so we are only considering values for $n$ that are positive. Notice that for values greater than $1$, $\frac{n^2-n-1}{n^4+n^2+1}$ can never be negative. $\endgroup$ Mar 16, 2021 at 23:19
  • $\begingroup$ So you are saying that becuase the only negative fraction of the sum is $-\frac{1}{3}$ then there exist elements that if we sum them we get a positive number bigger then $\frac{1}{3}$? $\endgroup$
    – Chopin
    Mar 16, 2021 at 23:24
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    $\begingroup$ Thank you a lot! $\endgroup$
    – Chopin
    Mar 16, 2021 at 23:29

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