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$$\sum_{n=1}^\infty \frac{4n}{n^4+2n+9}$$ Hi guys! I need to use comparison test on this series, I haven't done a lot of comparison tests so far, so I'm not sure what to compare it with. Should I use $$\frac{4n}{n^4}$$ which is $$\frac{4}{n^3},$$ but then I have to show convergence of $$\frac{4}{n^3}.$$ Do I keep comparing until I get to $$\frac{1}{n^2}$$ which I know and shown its convergence at a previous exercise? If anyone can help me with it, I would greatly appreciate it. Thank you!

metamorphy
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Stefana
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    Sounds like a good plan to me. Depending on the text you are using, it may have happened that the convergence of $\sum_n\dfrac1{n^3}$ was proven at the same time as that of $\sum_n\dfrac1{n^2}$. Using that would save you a bit of work, but the comparison also works in your favor. – Jyrki Lahtonen Nov 01 '19 at 10:13
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    TeX-Tip: Leave the punctuation marks inside the double-dollars. Otherwise the end up at the beginning of the next line, which looks strange. – Jyrki Lahtonen Nov 01 '19 at 10:15
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    See how the denominator can be increased, specifically $\frac{4n}{n^4+2n+9}\leq\frac{4n}{n^4+2n}\leq\frac{4n}{n^4}$ and then your logic works. It's always useful to think this way, rather than thinking what one can use to compare. Is this useful? – Anirudh Bhashyam Nov 01 '19 at 10:31
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    $n \gt 4:$ $0<4/n^3 <n/n^3 =1/n^2.$. – Peter Szilas Nov 01 '19 at 10:35
  • @Jyrki Lahtonen I don't have the convergence of 1/n^3, but I can prove it. Thanks! – Stefana Nov 01 '19 at 10:44
  • @Anirudh Bhashyam it is, thank you! – Stefana Nov 01 '19 at 10:44

2 Answers2

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More simply use limit comparison test with $\sum \frac1{n^2}$ and since

$$\frac{\frac{4n}{n^4+2n+9}}{\frac1{n^2}}= \frac{4n^3}{n^4+2n+9}\to0$$

we conclude that the series converges.

user
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First we try to bound the series above:

\begin{align} \sum_{n=1}^N\frac{1}{n^2} &=\sum_{n=2}^N\frac{1}{n^2}\\ &<1+\sum_{n=2}^N\frac{1}{n(n-1)} \\ &=1+\sum_{n=2}^N(\frac{1}{n-1}-\frac{1}{n})\\ &=1+1+\frac{1}{N}\overset{N\to\infty}{=}2 \end{align}

By comparison test, since $$\sum_{n=1}^\infty\frac{1}{n^3}\le\sum_{n=1}^\infty\frac{1}{n^2}\le2$$

The following series convers:

$$\sum _{n=1}^{\infty}\frac{1}{n^3}\tag*{$\text{method (1)}$}$$

By p-series test, since $p=3>1$, the following series converges:

$$\sum _{n=1}^{\infty}\frac{1}{n^3}\tag*{$\text{method (2)}$}$$

By integral test, since $\int_1^{\infty}\frac{1}{x^3}dx=\frac{1}{2}$ converges, imples the following series conveges:

$$\sum _{n=1}^{\infty \:}\frac{1}{n^3}\tag*{$\text{method (3)}$}$$

Since $n^4+2n+9>n^4$, then we have

$$\sum _{n=1}^{\infty \:}\frac{n}{n^4+2n+9}\le \sum _{n=1}^{\infty \:}\frac{n}{n^4}=\sum _{n=1}^{\infty \:}\frac{1}{n^3}$$

By comparison test, the following series convers:

$$\sum _{n=1}^{\infty \:}\frac{n}{n^4+2n+9}=L\text{, where } L\in\mathbb{R}$$

Which implies:

$$\sum _{n=1}^{\infty \:}\frac{4n}{n^4+2n+9}=4L\in\mathbb{R}$$

Therefore $\sum _{n=1}^{\infty \:}\frac{4n}{n^4+2n+9}$ converges

Ethan
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