# Convergence of Series (Harder)

1. Apr 29, 2012

### sid9221

Prove that:
$$(1-\frac{1}{n})^n \rightarrow \frac{1}{e}$$ as $$n \to \infty$$

you may use the fact that

$$(1+\frac{1}{n})^n \rightarrow e$$

I have no idea where to even begin, can someone point me in the right direction ?

2. Apr 29, 2012

### sharks

Try this: Multiply both sides by e.
$$(1-\frac{1}{n})^n .e \rightarrow \frac{1}{e} .e$$
$$(1-\frac{1}{n})^n .(1+\frac{1}{n})^n \rightarrow 1$$
If you can simplify the L.H.S. to get 1 on the R.H.S., then you have proved the convergence of the series.

3. Apr 29, 2012

### sid9221

Really don't know if you can sub in a sequence like that besides I need to prove it converges to $$\frac{1}{e}$$ not that it simply converges.

4. Apr 29, 2012

### sharks

$$(1-\frac{1}{n})^n .(1+\frac{1}{n})^n \rightarrow 1$$
$$( (1-\frac{1}{n}).(1+\frac{1}{n})) ^n \rightarrow 1$$
$$(1-\frac{1}{n}+\frac{1}{n}-\frac{1}{n^2})^n \rightarrow 1$$
$$(1-\frac{1}{n^2})^n \rightarrow 1$$
$$\lim_{n\to \infty}(1-\frac{1}{n^2})^n=1$$

5. Apr 29, 2012

### vela

Staff Emeritus
Let $a_n = \left(1 - \frac{1}{n}\right)^n$. Try determining what $\log a_n$ converges to.