Adsolute and conditional convergence of alternating series

  • Thread starter blursotong
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  • #1
blursotong
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i have a question regarding adsolute and conditional convergence of alternating series.

- i know that summation of [ tan(pi/n) ] diverge, but how do we proof it converge conditionally? (ie, (-1)^n tan(pi/n) ]

can Leibiniz's theorem be used in this case? but tan(pi/2) is infinite?

any help is appreciated. =D
 

Answers and Replies

  • #2
Dick
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If you write it as sum n=3 to infinity then you can use the alternating series test. If the series includes n=2 then it would be undefined.
 
  • #3
blursotong
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so alternating series of tan(pi/n) converge conditionally for n>3 only ?
for n>0 it is diverge?
 
  • #4
dacruick
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Conditional convergence of an alternating series means that it converges but if you take the absolute value it diverges?
 
  • #5
Dick
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so alternating series of tan(pi/n) converge conditionally for n>3 only ?
for n>0 it is diverge?

Maybe. Read the fine print in the definition and consult a lawyer. I would prefer to call the case n>0 undefined rather than divergent.
 
  • #6
Dick
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Conditional convergence of an alternating series means that it converges but if you take the absolute value it diverges?

Well, yes.
 

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