Proving convergence of factorial w/o Ratio Test

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Homework Help Overview

The discussion centers around determining the convergence or divergence of the series 1/n!. Participants are exploring methods within the context of comparison tests, as they have not yet covered the ratio test.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster considers using limit comparison with the series b_n=1/n^n but expresses uncertainty about its convergence. Other participants question the choice of b_n and suggest using simpler series for comparison. There is a discussion about the properties of n! and its relationship to n^2 and 2^n.

Discussion Status

Participants are actively engaging with the problem, offering various insights and suggestions for approaches. Some guidance has been provided regarding the use of the comparison test and the properties of the factorial function, but no consensus has been reached on a definitive method.

Contextual Notes

Participants are limited to using the comparison tests, p-series, geometric series, divergence test, and integral test, as these are the only methods they have learned so far.

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Homework Statement


Determine whether 1/n! diverges or converges.
So far, we have only learned the comparison tests, p-series, geometric series, divergence test, and integral test, so I can only use these tests to prove it.


Homework Equations



N/a

The Attempt at a Solution



I thought about using limit comparison with my b_n=1/n^n, but I can't determine if that converges or not, so I don't know what to do.
 
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Why did you choose b_n = 1/n^n? Why not a simpler series that you know converges or diverges?
 
The Comparison Test is your friend
 
It should be clear that for n> 3, n^2> n!.
 
HallsofIvy said:
It should be clear that for n> 3, n^2> n!.

I think you meant 2^n < n! ? Easy typo to make.

Oops, it's 2^(n-1) < n!
 
It's monotonic, so if you show that it is bounded, you're in business. Now by the comparison test, we have n!>n^2 for sufficiently large n, so take reciprocals and go from there.
 

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