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

1. Feb 25, 2009

### walker242

1. The problem statement, all variables and given/known data
Calculate the limit of $$a_{n} = \frac{n^n}{(n-1)^n}$$ when $$n\rightarrow \infty$$, where n is an integer.

2. Relevant equations
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3. 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?

2. Feb 26, 2009

### 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.

3. Feb 26, 2009

### Staff: Mentor

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:
$$\lim_{n \rightarrow \infty} \frac{ln(1 - 1/n}{1/n})$$

4. Feb 26, 2009

### walker242

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.