Sequence (n)/(n^n) Convergent or Divergent and Limit?

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SUMMARY

The sequence defined by {(n!)/(n^n)} is convergent, with its limit approaching zero as n approaches infinity. The analysis involves comparing the sequence to the ratio of consecutive terms, specifically examining the behavior of the terms as n increases. The ratio test confirms that the sequence converges since the growth of n! is outpaced by n^n. The conclusion is that the limit of the sequence is indeed zero.

PREREQUISITES
  • Understanding of factorial notation and properties (n!)
  • Knowledge of limits and convergence in sequences
  • Familiarity with the ratio test for convergence
  • Basic algebraic manipulation of sequences and inequalities
NEXT STEPS
  • Study the Ratio Test for convergence in more detail
  • Explore Stirling's approximation for factorials
  • Learn about the comparison test for sequences
  • Investigate the behavior of sequences involving exponential growth
USEFUL FOR

Students studying calculus, particularly those focusing on sequences and series, as well as educators looking for examples of convergence in mathematical analysis.

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Homework Statement


Is the sequence {(n!)/(n^n)} convergent or divergent. If it is convergent, find its limit.

Homework Equations


Usually with sequences, you just take the limit and if the limit isn't infinity, it converges... That doesn't really work here. I know I'm supposed to write out the terms, but I can't figure out where to go from there.

The Attempt at a Solution


(n!)/(n^n) = ((1)(2)(3)...(n-1)(n))/((n)(n)(n)...(n)(n))
I think I'm supposed to say that this is less than something that goes to zero, but I don't know what that something should be.

n/n = 1

Does this mean that I can cancel that out and have a list of numbers smaller than n divided by n? Then the sequence is less than 1...but I think it converges to zero, I just don't know how to show that.
 
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You are on the right track. Try writing out a few terms to see if you can see the growth factor from ##\frac{n!}{n^n}## to ##\frac{(n+1)!}{(n+1)^{(n+1)}}##.
 
Okay, so...

((n+1)!)/((n+1)^(n+1)) = (1/(n+1)) ((n+1)!)/((n+1)^n) = (1/(n+1)) (((1)(2)(3)...(n)(n+1))/((n+1)(n+1)(n+1)...(n+1)(n+1)) < (1/(n+1)) and the limit of 1/(n+1) as n goes to infinity is zero...

Is my inequality true? My logic was just that the huge set of terms being multiplied by each other is still less than 1, so a (1/(n+1)) is going to be greater than itself multiplied by a number less than one. Is this correct?

How exactly does this relate to n!/(n^n)?
 
RUber is suggesting you consider the ratio between consecutive terms, \frac {n!}{n^n} and \frac {(n+1)!}{(n+1)^{n+1}}.
 
The ratio test would work I believe. In general though: ##n^n > n! > b^n > n^x > \ln(n)## as ##n \rightarrow \infty##. So its easy to determine convergence by merely examining the behaviour of ##a_n##.
 
I think you have said it for the n+1 case. The logic holds for the nth case just the same.
##\frac{n!}{n^n}=\frac1n \frac2n ... \frac nn ## the largest term in the product is 1, so the product can be no larger than the smallest term. Just as you said in post #3.
 

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