Does the Nth Term Test Indicate Divergence for \(\sum (1-\frac{3}{n})^n\)?

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SUMMARY

The discussion centers on the convergence of the series \(\sum (1-\frac{3}{n})^n\) and its relation to the Nth Term Test. The Nth Term Test confirms that this series diverges because the limit of the terms approaches \(e^{-3}\), which is not zero. This contrasts with the theorem stating that \(\sum (1+\frac{x}{n})^n\) converges to \(e^x\), highlighting a distinction between the convergence of the series and the limit of its terms.

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kuahji
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For a problem such as [tex]\sum[/tex] (1-3/n)^n with the sum going from 1 to infinity, the textbook shows to use the nth term test & that it diverges. However in a previous chapter the textbook had a theorem that said the limit of [tex]\sum[/tex] (1+x/n)^n will converge to e^x for any number x.

So I'm confused. Are they using convergence in a different context or what am I missing?
 
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It is (1+x/n)^n that converges to e^x, not the sum of that.

And actually, you can use this fact to show that the series in your problem diverges. You know that (1-3/n)^n converges to e^{-3}, which is not 0, hence the series diverges.
 
Last edited:
Thanks for the reply. I see that I obviously got confused ^_^.
 

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