Convergence/divergence of a series

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In summary, the original question asks for the radius and interval or convergence, and the values of x for which the series \sum_{n=1}^{\infty}(1+\frac{1}{n})^{n}x^n will converge absolutely and conditionally. I have 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. So I am trying to show whether \sum_{n=1}^{\infty}(-1)^{n}(1+\frac{1}{n})^{n} conver
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miglo
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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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  • #2
All that stuff does not matter. A series can only converge if lim an=0, so this series cannot converge.
 
  • #3
i thought it would something really simple, i guess i was just over thinking it
thanks!
 

1. What is the definition of convergence/divergence of a series?

The convergence or divergence of a series refers to whether the sum of all the terms in a sequence approaches a finite value (convergence) or approaches infinity (divergence) as the number of terms increases.

2. How do you determine if a series is convergent or divergent?

To determine the convergence or divergence of a series, you can use various tests such as the ratio test, comparison test, or integral test. These tests involve evaluating the behavior of the terms in the series and determining if they approach a finite value or infinity.

3. What is the difference between absolute and conditional convergence?

Absolute convergence refers to a series where the sum of the absolute values of the terms converges, regardless of the order of the terms. Conditional convergence, on the other hand, refers to a series where the sum of the terms converges only when the terms are arranged in a specific order.

4. Can a series be both convergent and divergent?

No, a series can only be either convergent or divergent. If the sum of the terms in a series approaches a finite value, it is considered to be convergent. If the sum of the terms approaches infinity, it is considered to be divergent.

5. What is the importance of determining the convergence or divergence of a series?

Determining the convergence or divergence of a series is important in various mathematical applications, such as in calculating the sums of infinite sequences or in solving differential equations. It also helps in understanding the behavior of a series and determining the accuracy of mathematical models that involve infinite series.

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