Proof that series diverges ?

  • Thread starter sid9221
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  • #1
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[tex] \sum_{1}^{\infty} (-1)^n(1+\frac{1}{n})^n [/tex]

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

Can't think of any other test...
 

Answers and Replies

  • #2
tiny-tim
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hi sid9221! :smile:

hint: draw an against n on graph paper (as dots) …

what does it look like? :wink:
 
  • #3
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I know I've seen the plot somewhere (can't remember the technical name).

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

I know it diverges just can't prove it.
 
  • #4
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Use the "n-th term test for divergence." If the limit of the n-th term in the series is different from 0 or doesn't exist, the series diverges.
 
  • #5
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How to you take the limit of (-1)^n.

Wolfram alpha says e^2i 0 to Pi !?!!
 
  • #6
Ray Vickson
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How to you take the limit of (-1)^n.

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

RGV
 

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