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Determine whether the series

$\sum _{n=1}^{\infty }\:\frac{n^2-5n}{\sqrt{n^7+2n+1}}$

is convergent or divergent.

So far in class i've learned a lot of different test to use, but i'm having trouble finding out what test would be 'most pratical' for this problem. The ratio test would be too complicated so i was thinking possibly the comparison test. usually i think this test is easy to see what it would compare to

$\:0\le \frac{n^2-5n}{\sqrt{n^7+2n+1}}\le \frac{n^2}{\sqrt{n^7}}$

this is what i would assume because i know i want to keep the 'key items' but at the same time i've never done a comparison test with a square root in the bottom.

Taking $\frac{n^2}{\sqrt{n^7}}$ =$\frac{n^2}{n^{\frac{7}{2}}}$ = $\frac{1}{n^{\frac{3}{2}}}$

$\:\sum _{n=1}^{\infty }\:\frac{1}{n^{\frac{3}{2}}}$ p-series with p = $\frac{3}{2}$ $\ge $ 1

Therefore Convergent

and by Comparison Test means

$\sum _{n=1}^{\infty }\:\frac{n^2-5n}{\sqrt{n^7+2n+1}}$ is also convergent

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  • $\begingroup$ thank you! and then it's convergent because its a p-series p=3/2 which is greater than 1! $\endgroup$ – Charlene Nov 14 '14 at 20:15
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One issue with your attempt to use Direct Comparison test arises from your inequality $$0\leq \frac{n^2-5n}{\sqrt{n^7+2n+1}}\leq \frac{n^2-5n}{\sqrt{n^7}}$$ because $n^2-5n$ is not always greater than or equal to $0$. Indeed, this is the case only when $n=0$ and $n\geq 5$. Of course, you can simply apply the Direct Comparison test to the series from $n=5$ to $\infty$, so this is not an irreparable issue.

For $n\geq 5$, we have $0\leq n^2-5n\leq n^2$. Thus, instead of using the series you did above for Direct comparison, why not do it with $\frac{n^2}{\sqrt{n^7}}=\frac{1}{n^{3/2}}$?

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Do a comparison test with $n^{-\frac{3}{2}}=\frac{n^2}{\sqrt{n^7}}>\frac{n^2-5n}{\sqrt{n^7}}>\frac{n^2-5n}{\sqrt{n^7+2n+1}}$

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