# Basic harmonic series (not sure)

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]

• 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. Commented Mar 24, 2021 at 18:39
• hint: For $n>17$ then $n+17< ?$ and $4n^2+5>0$. So $c=\frac 13$ works.
– zwim
Commented Mar 24, 2021 at 18:43

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 $$c = 6$$: