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The series I am presented with: $$\sum_{n=1}^{\infty}\frac{n+17}{6n^3+4n^2+5}$$ How can I find the series $\sum_{n=1}^{\infty}b_n$ of the form $\sum_{n=1}^{\infty}b_n=c\sum_{n=1}^{\infty}\frac{1}{n^2}$ where $c\gt0$ such that $b_n\ge\frac{n+17}{6n^3+4n^2+5}$ for all $n\in\mathbb{N}$? [Side note: I'm not entirely sure if this is in the Harmonic Series subject]

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    $\begingroup$ This is not the harmonic series: the harmonic series is $\sum \frac{1}{n}$. For your example, show that $\frac{\frac{n+17}{6n^3 + 4n^2 +5}}{\frac{1}{n²}}$ is bounded and you can now conclude. $\endgroup$
    – Didier
    Commented Mar 24, 2021 at 18:39
  • $\begingroup$ hint: For $n>17$ then $n+17< ?$ and $4n^2+5>0$. So $c=\frac 13$ works. $\endgroup$
    – zwim
    Commented Mar 24, 2021 at 18:43

1 Answer 1

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Simplifying the following inequality $$\begin{align} &\qquad \ \ \ \ \frac{n+17}{6n^3+4n^2+5} \le \frac{c}{n^2} \\ &\iff n^3+17n^2 \le 6cn^3+4cn^2+5c \\ &\iff(6c-1)n^3 + (4c-17)n^2 + 5c \ge 0 \end{align}$$ Taking $4c > 17$ or $c > 5$ is enough.


You can see how it works with $c = 6$:

enter image description here

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