Series converges or diverges

**tangibleLime** · December 14th 2010, 01:46 PM

Determine whether the series converges or diverges.

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.

$\sum_{n=1}^{\infty} \frac{n}{n^\frac{7}{3}} = \sum_{n=1}^{\infty} \frac{1}{n^\frac{4}{3}}$

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.

$\lim_{n \to \infty} | \frac{a_{n}}{b_{n}} |$

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

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

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?

**Plato** · December 14th 2010, 02:01 PM

$n \geqslant 2\, \Rightarrow \,\dfrac{{n + 2}}<br /> {{\sqrt[3]{{n^7 + n^2 }}}} \leqslant \dfrac{{2n}}<br /> {{\sqrt[3]{{n^7 }}}} = \dfrac{2}<br /> {{n^{\frac{4}<br /> {3}} }}$

**tangibleLime** · December 14th 2010, 02:16 PM

If I understand you correctly, you are comparing it to,

$\frac{2n}{\sqrt[3]{{n^7}}}$

, since the addition of the n^2 in the original denominator is negligible, as well as the addition of the 2 in the original numerator. Though I am a bit confused about where the 2n came from in the numerator in the comparison series? Could I just leave it as a 1 and use the p-series test to prove convergence?

**Plato** · December 14th 2010, 02:42 PM

If $n\ge 2$ then $n+2\le 2n$ . Simple?

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