Series againand again

  • Thread starter Alem2000
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  • #1
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Sorry about the title everyone but ive posted numerous threads on series and I had to choose an apropriate title :tongue2:
The problem asks to use the ratio test, and determine for which values of x the test is conclusive-either converging or diverging. Then check those cases where the test is inconclusive by some other means.

here is the the series [tex]\sum_{n=3}^{\infty}\frac{x^n}{n3^n}[/tex]...converge or diverge here is what i did [tex]\frac{a_{n+1}}{a_n}[/tex] and that came out to be [tex]\frac{x^{n+1}}{(n+1)(3^{n+1})}[/tex] multiplie by the [tex]\frac{n3^{n}}{x^{n}}[/tex] and after you cross out similar variables and it comes out to be

[tex]\lim_{x\rightarrow \infty}\frac{xn}{3(n+1)}[/tex]
 
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Answers and Replies

  • #2
Galileo
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[tex]\frac{a_{n+1}}{a_n}=\frac{x^{n+1}n3^n}{x^n(n+1)3^{n+1}}=\frac{nx}{(n+1)3}[/tex]
 
  • #3
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thanks galileo but I got that far just had problems latexing it
 
  • #4
So you get

[tex]\lim_{n \to \infty} \left| \frac{n}{(n+1)}\frac{x}{3}\right| = \frac{|x|}{3}[/tex]

And you know this is stricly less than 1 for it to be conclusively converging, strictly greater than 1 to be conclusively diverging, and inconclusive at 1. So you must test the values for which the expression equals one.

--J
 
  • #5
James R
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The remaining cases are x=3 and x=-3

If x=3 then the series reduces to the harmonic series, which diverges.

If x=-3 then we have an alternating series, and we can use the Leibnitz test (whose exact conditions escape me right now).
 

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