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## Homework Statement

Determine whether the series converges or diverges.

## Homework Equations

## The Attempt at a Solution

I got the correct answer but I'm not sure if my method is correct, or if I made so many mistakes that I ended up with the correct answer. If someone could go over this and tell me if my math is perfect or flawed, that would be fantastic.

First I set up a comparison series. The (7/3) comes from taking the n^7 out of the cubed root.

[tex]\sum_{n=1}^{\infty} \frac{n}{n^\frac{7}{3}} = \sum_{n=1}^{\infty} \frac{1}{n^\frac{4}{3}}[/tex]

This new series is a

*convergent*p-series since p = (4/3), and (4/3) > 1.

So to test if the original series is also convergent, I have to use either the direct comparison test or the limit test. The limit test looks like it would be a pain to do, so I tried the direct comparison test. Unfortunately, the original series is not less than or equal to the new series, so the direct comparison test would not work. So I used the limit test.

[tex]\lim_{n \to \infty} | \frac{a_{n}}{b_{n}} |[/tex]

[tex]\lim_{n \to \infty} | \frac{n^{\frac{4}{3}}*n+2}{\sqrt[3]{n^{7}+n^{2}}} |[/tex]

[tex]\lim_{n \to \infty} | \frac{n^{\frac{7}{3}}+2}{\sqrt[3]{n^{7}+n^{2}}} |[/tex]

Seeing that the highest power up top is equal to the highest power on bottom (7/3), I used to shortcut that the limit will end up being the ratio of those two coefficients (in this case, 1/1 = 1).

So b_n is >0, the limit is 1, which is a positive, finite number, which confirms that the original series is convergent.

I think the radical is throwing me off... I don't know what to do with it. When I try to take the limit do I divide all the terms by n^7, or n^7/3 since it's under the radical? Or is it fine the way I did it?

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