# How do you determine whether the sequence a_n=((n-1)/n)^n converges, if so how do you find the limit?

Feb 7, 2017

${\lim}_{n \to \infty} {a}_{n} = \frac{1}{e}$

#### Explanation:

We have:

${a}_{n} = {\left(\frac{n - 1}{n}\right)}^{n}$

this is a positive number so we can take the logarithm:

$\ln {a}_{n} = \ln {\left(\frac{n - 1}{n}\right)}^{n} = n \ln \left(\frac{n - 1}{n}\right)$

Now if:

${\lim}_{x \to \infty} x \ln \left(\frac{x - 1}{x}\right)$

exists is must be the same as:

${\lim}_{n \to \infty} \ln {a}_{n} = {\lim}_{n \to \infty} n \ln \left(\frac{n - 1}{n}\right)$

This limit in the indeterminate form $0 \cdot \infty$ so we have to transform it to apply l'Hospital's rule:

${\lim}_{x \to \infty} x \ln \left(\frac{x - 1}{x}\right) = {\lim}_{x \to \infty} \frac{\ln \left(\frac{x - 1}{x}\right)}{\frac{1}{x}} = {\lim}_{x \to \infty} \frac{\frac{d}{\mathrm{dx}} \ln \left(\frac{x - 1}{x}\right)}{\frac{d}{\mathrm{dx}} \frac{1}{x}}$

Now:

$\frac{d}{\mathrm{dx}} \ln \left(\frac{x - 1}{x}\right) = \frac{d}{\mathrm{dx}} \left(\ln \left(x - 1\right) - \ln x\right) = \frac{1}{x - 1} - \frac{1}{x} = \frac{1}{x \left(x - 1\right)}$

$\frac{d}{\mathrm{dx}} \frac{1}{x} = - \frac{1}{x} ^ 2$

so:

${\lim}_{x \to \infty} x \ln \left(\frac{x - 1}{x}\right) = {\lim}_{x \to \infty} \frac{\frac{1}{x \left(x - 1\right)}}{- \frac{1}{x} ^ 2} = {\lim}_{x \to \infty} - {x}^{2} / \left(x \left(x - 1\right)\right) = - 1$

and we can conclude that:

${\lim}_{n \to \infty} \ln {a}_{n} = - 1$

But since $\ln x$ is a continuous function:

${\lim}_{n \to \infty} \ln {a}_{n} = \ln \left({\lim}_{n \to \infty} {a}_{n}\right) = - 1$

which implies:

${\lim}_{n \to \infty} {a}_{n} = \frac{1}{e}$