Series Comparison Test for (n^n)/n! and Convergence Analysis

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Homework Help Overview

The discussion revolves around determining the convergence or divergence of the series given by the sum from n = 1 to infinity of (n^n)/n!. Participants are exploring the application of the comparison test and the ratio test in this context.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to apply the comparison test and are questioning the cancellation of terms in the limit as n approaches infinity. There is also a suggestion to consider the ratio test as an alternative approach.

Discussion Status

The discussion is ongoing, with participants providing insights and questioning each other's reasoning. Some guidance has been offered regarding the manipulation of terms, but there is no explicit consensus on the approach or outcome yet.

Contextual Notes

Participants are navigating through the implications of their assumptions about term cancellation and the behavior of the series as n becomes large. There is an acknowledgment of differing interpretations of the expressions involved.

phrygian
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Homework Statement



Find if the sum from n = 1 to infinity of (n^n)/n! diverges or not.

Homework Equations



p = an+1/an


The Attempt at a Solution



Using the comparison test I get to the point where p_n = (n+1)^(n+1) / [(n+1) n^n]

Shouldnt p just be 0, don't (n+1)^(n+1) and n^n cancel for large n? The book says the answer is p = e, how do you get there?

Thanks for the help!
 
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Why not try the ratio test?
 
nn+1/nn doesn't cancel for large n, so when you replace nn+1 with (n+1)n+1 why would you expect it to?
 
No, they don't cancel for large n. You are jumping to conclusions. If I look at your expression I would write it as (n+1)^n/n^n. Do you see how? Now what?
 
not zero, cancel the n+1 term
[tex]\frac{(n+1)^{n+1}}{(n+1) n^n} = \frac{(n+1)^n}{n^n} = (\frac{n+1}{n})^n[/tex]
 

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