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Calculate the limit of $a_{n} = \frac{n^n}{(n-1)^n}$ when $n\rightarrow\infty$

  • Thread starter walker242
  • Start date
  • #1
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Homework Statement


Calculate the limit of [tex]a_{n} = \frac{n^n}{(n-1)^n}[/tex] when [tex]n\rightarrow \infty[/tex], where n is an integer.


Homework Equations


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The Attempt at a Solution


[tex]a_{n} = \frac{n^n}{(n-1)^n} = \left(\frac{n}{n-1}\right)^{n} = e^{\ln\frac{n}{n-1}\right)^{n}} = e^{n\ln\frac{n}{n-1}} = e^{n\left(\ln(n)-\ln(n-1)\right)} = e^{-n\left(\ln(n-1)-\ln(n)\right)} = e^{-n\left(\ln\frac{n-1}{n}\right)} = e^{-n\left(\ln\left(1-\frac{1}{n}\right)\right)} = [/tex]que?
 

Answers and Replies

  • #2
CompuChip
Science Advisor
Homework Helper
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What is the limit of
[tex]n \ln(1 - 1/n)[/tex]
as [itex]n \to \infty[/itex]?
When you're not sure, try expanding the logarithm in a series around n = infinity.
 
  • #3
33,478
5,168
For this limit--
[tex]\lim_{n \rightarrow \infty} n ln(1 - 1/n)[/tex]
a simpler approach might be to use L'Hopital's Rule, with the limit rewritten as a quotient:
[tex]\lim_{n \rightarrow \infty} \frac{ln(1 - 1/n}{1/n})[/tex]
 
  • #4
12
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While both suggestions do give me the end result I want, we have still not reached MacLaurin expansions, nor L'Hopitals rule. Much thanks for your help, however.
 

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