Convergence/divergence of a series

  • Thread starter miglo
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  • #1
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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 im 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 doesnt 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] isnt decreasing nor will the limit as n approaches infinity go to 0, so i cant use the alternating series test to show whether its converging or diverging
so what do i do now? can i just say since it doesnt satisfy the requirements necessary to make it convergent by the alternating series test then it will diverge? or is there some other method im not seeing?
 

Answers and Replies

  • #2
lurflurf
Homework Helper
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All that stuff does not matter. A series can only converge if lim an=0, so this series cannot converge.
 
  • #3
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i thought it would something really simple, i guess i was just over thinking it
thanks!
 

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