Classifying an Alternating Factorial Series

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Homework Help Overview

The problem involves classifying the convergence of the series \(\Sigma^{\infty}_{k = 1} (-1)^{k-1}\frac{k!}{(2k-1)!}\), specifically determining whether it is absolutely convergent, conditionally convergent, or divergent. The context is within the study of alternating series and factorials.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the Alternating Series Test and express uncertainty about determining if the series is decreasing due to the factorial nature of the terms. There are attempts to apply the ratio test to analyze the terms, but some participants encounter difficulties with the factorial expressions.

Discussion Status

The discussion reflects a collaborative exploration of the problem, with participants offering suggestions on using the Alternating Series Test and the ratio test. There is acknowledgment of challenges faced with factorials, and one participant indicates they were able to solve the problem, suggesting some progress has been made.

Contextual Notes

Participants are navigating the complexities of factorials in the context of convergence tests, which may introduce assumptions about the behavior of the series that are still under discussion.

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Homework Statement


Classify the series as absolute convergent, conditionally convergent, or divergent.

[tex] \Sigma^{\infty}_{k = 1} (-1)^{k-1}\frac{k!}{(2k-1)!}[/tex]


Homework Equations


The Alternating Series Test: conditions for convergence
decreasing
lim --> infinity ak = 0


The Attempt at a Solution


I am not sure how to find out if the series is decreasing. Since it is a factorial, I cannot take the first deriative test.

I was wondering if I should use the Alternating Series Test at all, since it is an factorial.

Can someone help me here?
 
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try the alternating series test

a ratio of terms will show if the magnitude of terms decrease monotonically
 
Thanks.

If I try the ratio I got stuck with the factorial.

[tex] \frac{(k+1)!}{(2(k+1)-1)!} * \frac{(2k-1)!}{k!} = \frac{(2k-1)!}{(2(k+1)-1)!} * (k+1)[/tex]


Can I write

[tex](2(k+1)-1)! = (2k+2-1)! = (2k+1)![/tex]
 
dba said:
Thanks.

If I try the ratio I got stuck with the factorial.

[tex] \frac{(k+1)!}{(2(k+1)-1)!} * \frac{(2k-1)!}{k!} = \frac{(2k-1)!}{(2(k+1)-1)!} * (k+1)[/tex]


Can I write

[tex](2(k+1)-1)! = (2k+2-1)! = (2k+1)![/tex]
Yes, of course. And so you have (2k-1)!/(2k+1)!= (2k-1)!/[(2k+1)(2k)(2k-1)!]
 
Oh, ok.
Thank you! I was able to solve this one :smile:
 

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