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

[walker242]

Homework Statement

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

Homework Equations

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The Attempt at a Solution

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)} =que?

 

[CompuChip]

What is the limit of
n \ln(1 - 1/n)
as n \to \infty?
When you're not sure, try expanding the logarithm in a series around n = infinity.

 

[Mark44]

For this limit--
\lim_{n \rightarrow \infty} n ln(1 - 1/n)
a simpler approach might be to use L'Hopital's Rule, with the limit rewritten as a quotient:
\lim_{n \rightarrow \infty} \frac{ln(1 - 1/n}{1/n})

 

[walker242]

While both suggestions do give me the end result I want, we have still not reached MacLaurin expansions, nor l'hospital's rule. Much thanks for your help, however.