- #1

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- 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

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- Thread starter blursotong
- Start date

- #1

- 15

- 0

- 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

- #2

Dick

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- #3

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so alternating series of tan(pi/n) converge conditionally for n>3 only ?

for n>0 it is diverge?

for n>0 it is diverge?

- #4

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- #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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Well, yes.

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