Does the Series \(\sum_{n=1}^{\infty} (-1)^n(1+\frac{1}{n})^n\) Diverge?

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Homework Help Overview

The discussion centers around the convergence or divergence of the series \(\sum_{n=1}^{\infty} (-1)^n(1+\frac{1}{n})^n\), with participants exploring various tests and approaches to analyze the series behavior.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the alternating series test and the n-th term test for divergence, questioning the behavior of the terms as \(n\) approaches infinity. There are attempts to visualize the series through graphing and references to its resemblance to a sine function.

Discussion Status

The discussion is ongoing, with participants offering hints and exploring different interpretations of the series' behavior. Some guidance has been provided regarding the n-th term test, but no consensus has been reached on the series' convergence or divergence.

Contextual Notes

There are mentions of potential confusion regarding the limit of the alternating term and the implications of its behavior on the overall series. Participants are also considering the significance of the terms approaching zero in the context of divergence tests.

sid9221
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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...
 
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hi sid9221! :smile:

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

what does it look like? :wink:
 
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.
 
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.
 
How to you take the limit of (-1)^n.

Wolfram alpha says e^2i 0 to Pi ?!
 
sid9221 said:
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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