Series converges or diverges

  • #1

Homework Statement



Determine whether the series converges or diverges.

symimage.cgi?expr=sum_%28n%3D1%29%5E%28infinity%29%20%28n%2B2%29%2Froot3%28n%5E7%2Bn%5E2%29.gif


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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Answers and Replies

  • #2
867
0
Pull out n7/3 from the radical, simplify the expression a bit by making the n7/3's cancel, and then take the limit.
 
  • #3
Well that's what I'm having difficulty with... pulling terms out of that radical. I've searched the Internet high and low and apparently there is no example of pulling out a term of a radical that is being added.

For example:

[tex]\sqrt{4^{2}+5^{2}}[/tex]

[tex](4^{2}+5^{2})^{1/2}[/tex]

[tex]4^{2/2}+5^{2/2}[/tex]

[tex]4^{1}+5^{1}[/tex]

[tex]4+5[/tex]

[tex]9[/tex]

And I know that the original definitely does not equal 9, so this can't be a correct method... So I don't really know how I can just pull that n^7 out of that cubed root.
 
  • #4
1,101
3
What you reasoned in the original post should be fine. As n grows large, the n^2 term under the radical is insignificant, so the denominator behaves as n^(7/3) and it is clear the ratio a_n / b_n tends to 1. Don't forget to multiply the 2 in the numerator by n^(4/3).
 
  • #5
PAllen
Science Advisor
8,284
1,547
You can also use the comparison test. Just multiply your simple series by e.g. 2. Then for all n greater than some value, the augmented simple series will be larger. This is a common trick you should know.
 
  • #6
Ahh, makes sense. Thank you all!
 
  • #7
Reply to the algebra question: [itex]\sqrt{4^2 + 5^2}[/itex]

[itex]\sqrt{4^2 (1+ \frac{5^2}{4^2})}[/itex]

[itex][4^{2}(1+\frac{5^2}{4^2})]^\frac{1}{2}[/itex]

and by the law of exponents

[itex](a^{m}b^{n})^c = a^{mc}b^{nc} [/itex]

[itex]4^{\frac{2}{2}}\sqrt{1+ \frac{5^2}{4^2}}[/itex]

Thats it.
OMG I hate writing code, but this is to help others so smile!
 
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