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I've tried the alternating series test but the "b_n" part converges to e.

Can't think of any other test...

- Thread starter sid9221
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- #1

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I've tried the alternating series test but the "b_n" part converges to e.

Can't think of any other test...

- #2

tiny-tim

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hi sid9221!

hint: draw a_{n} against n on graph paper (as dots) …

hint: draw a

what does it look like?

- #3

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It looks like a sin function but one that's increasing .

I know it diverges just can't prove it.

- #4

Mark44

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How to you take the limit of (-1)^n.

Wolfram alpha says e^2i 0 to Pi !?!!

Wolfram alpha says e^2i 0 to Pi !?!!

- #6

Ray Vickson

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You are missing the whole point. If you have a series [itex] S = t_1 - t_2 + t_3 - t_4 + \cdots,[/itex] with all [itex] t_i > 0 \;( \text{or all } < 0),[/itex] you need [itex]|t_n| \rightarrow 0 [/itex] as [itex] n \rightarrow 0.[/itex] Of course the factors [itex] (-1)^n[/itex] do not have a limit, but that is not important.How to you take the limit of (-1)^n.

Wolfram alpha says e^2i 0 to Pi !?!!

RGV

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