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The series is $$\sum_{n=0}^\infty \frac{2n^2-1}{3n^5+2n+1}$$

I used the comparison method to solve which gave me $b_n=\frac{2n^2}{3n^5+2n}$, then I put those two into a limit with $\frac{a_n}{b_n}$

$$\lim_{n\to\infty}\frac{\frac{2n^2-1}{3n^5+2n+1}}{\frac{2n^2}{3n^5+2n}}=\lim_{n\to\infty}\frac{2n^2-1}{3n^5+2n-1}*\frac{3n^5+2n}{2n^2}=$$

$$\lim_{n\to\infty}\frac{2n^2-1\left(3n^5+2n\right)}{3n^5+2n+1\left(2n^2\right)}$$ After canceling out terms I'm left with $\lim_{n\to\infty}-1$ which means the series diverges but the answer says it should converge. Did I mess up with the cross multiplication or somewhere else?

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  • $\begingroup$ "which means the series diverges" Why is that? $\sum_n b_n$ converges. $\endgroup$
    – Clement C.
    Dec 3 '19 at 19:34
  • $\begingroup$ You can use L’Hopital’s rule to show the series converges. Ignoring all terms with degree lower than the highest degree on both the numerator and denominator , we get 2n^2/3n^5=2/3n^3 so the series converges. $\endgroup$
    – Jazzowner
    Dec 3 '19 at 19:37
  • $\begingroup$ @Jazzowner Come on, you don't need L'Hopital and the differentiations it requires to solve that one... $\endgroup$
    – Clement C.
    Dec 3 '19 at 19:39
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There is a serious error in your computation. Note that\begin{align}\lim_{n\to\infty}\frac{2n^2-1}{3n^5+2n-1}\cdot\frac{3n^5+2n}{2n^2}&=\lim_{n\to\infty}\frac{6n^7+\text{terms of lower degree}}{6n^7+\text{terms of lower degree}}\\&=1.\end{align}Therefore, the series converges.

Note that the limit could not possibly be negative, since each $a_n$ and each $b_n$ is positive.

On the other hand, it is much more natural to take $b_n=\dfrac1{n^3}$ instead.

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  • $\begingroup$ There is an error, but even at a high level: given the choice of $b_n$, even if the limit had indeed been $-1$ the conclusion would have been convergence. That is some deeper mistake from the OP's than just a computation one, I would say. $\endgroup$
    – Clement C.
    Dec 3 '19 at 19:37
  • $\begingroup$ Yes, you are right. $\endgroup$ Dec 3 '19 at 19:39
  • $\begingroup$ @JoséCarlosSantos so the final fraction should be $\frac{6n^7-3n^5+4n^3-2n}{6n^7+4n^3+2n^2}$? I know that doesn't equal 1 so I'm just wondering how I keep messing up. $\endgroup$
    – Eric Brown
    Dec 3 '19 at 19:45
  • $\begingroup$ @EricBrown Yes, that's what the final fraction looks like. $\endgroup$ Dec 3 '19 at 19:49
  • $\begingroup$ @JoséCarlosSantos so then how does it equal 1 if I still have a $-3n^5$ in the numerator and a $2n$ in the denominator? $\endgroup$
    – Eric Brown
    Dec 3 '19 at 19:57
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The series converges. First show that for all $n$ sufficiently large, $a_n\geq 0$. Then show for all $n$ sufficiently large, $a_n\leq \frac{1}{n^2}$. Now you can use comparison test.

By "sufficiently large", $n\geq 10$ will do just fine.

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

$$\sim \frac{2}{3n^\color{red}{3}}$$

$\color{red}{3}>1$ so, it is a positive convergent series.

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Step by step $$0\leq \sum_{n=0}^\infty \frac{2n^2-1}{3n^5+2n+1}\leq\sum_{n=0}^\infty \frac{2n^2}{3n^5+2n+1}\\\leq\sum_{n=0}^\infty \frac{2n^2}{3n^5+2n}\leq\sum_{n=0}^\infty \frac{2n^2}{3n^5}=\frac23\sum_{n=0}^\infty \frac{1}{n^3}$$so $$0\leq \sum_{n=0}^\infty \frac{2n^2-1}{3n^5+2n+1}\leq \frac23\sum_{n=0}^\infty \frac{1}{n^3}$$

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Apparently, you forgot to uses parentheses where required in $\frac{2n^2-1\left(3n^5+2n\right)}{3n^5+2n+1\left(2n^2\right)}$, which should be $\frac{\color{red}(2n^2-1\color{red})\left((3n^5+2n\right)}{\color{red}(3n^5+2n+1\color{red})\left(2n^2\right)}$.

This being will be simpler using symptotic equivalents: near $\infty$ a polynomial is equivalent to its leading term, and equivalence is compatible with multiplication and division, so $$ \frac{2n^2-1}{3n^5+2n+1}\sim_{n\to\infty}\frac{2n^2}{3n^5}=\frac23\frac1{n^3},$$ and the latter is the general term of a convergent $p$-series.

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