Convergence/divergence of a series

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SUMMARY

The discussion focuses on determining the convergence of the series \(\sum_{n=1}^{\infty}(1+\frac{1}{n})^{n}x^n\). The radius of convergence is established as \(R=1\) with the interval of convergence being \(-1 PREREQUISITES

  • Understanding of series convergence tests, specifically the absolute convergence test and the alternating series test.
  • Familiarity with limits and the behavior of sequences as \(n\) approaches infinity.
  • Knowledge of the properties of exponential functions, particularly \(e^x\).
  • Basic calculus concepts related to series and convergence intervals.
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Students and educators in calculus, particularly those focusing on series convergence, as well as mathematicians interested in advanced series analysis techniques.

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Homework Statement


the original question asks for the radius and interval or convergence, and the values of x for which the series [tex]\sum_{n=1}^{\infty}(1+\frac{1}{n})^{n}x^n[/tex] will converge absolutely and conditionally
ive figured out the radius of convergence (R=1) and my interval of convergence (-1<x<1) but now i need to check the endpoints of the interval of convergence to see where it converges absolutely and conditionally if it does in fact converge conditionally
so I am trying to show whether [tex]\sum_{n=1}^{\infty}(-1)^{n}(1+\frac{1}{n})^{n}[/tex] converges or diverges


Homework Equations


absolute convergence test
alternating series test


The Attempt at a Solution


so i first tried using the absolute convergence test which meant that i only had to look at the series without the (-1)^n, and then i applied the divergence test or n-th term test, taking the limit of the sequence as n approached infinity but that just gives me e^1 which is greater than 0 therefore it diverges but this doesn't imply that the original series with the (-1)^n attached to it will diverge so i have to use the alternating series test
but the sequence [itex](1+\frac{1}{n})^{n}[/itex] isn't decreasing nor will the limit as n approaches infinity go to 0, so i can't use the alternating series test to show whether its converging or diverging
so what do i do now? can i just say since it doesn't satisfy the requirements necessary to make it convergent by the alternating series test then it will diverge? or is there some other method I am not seeing?
 
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All that stuff does not matter. A series can only converge if lim an=0, so this series cannot converge.
 
i thought it would something really simple, i guess i was just over thinking it
thanks!
 

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