# Quick question on real analysis proof

1. Nov 3, 2013

### Lee33

1. The problem statement, all variables and given/known data

Show that the sequence of functions $x(1-x), x^2(1-x),...$ converges uniformly on $[0,1].$

2. The attempt at a solution

I have a quick question. For the following proof why is $\left ( \frac{n}{n+1}\right )^n < 1$?

Proof:

We need to prove that, given $\epsilon > 0$, there exists an $N$ such that for every $n > N$ and for every $x \in[0, 1]$, we have $|x^n(1 - x)-0| < \epsilon.$

$x^n$ and $(1-x)$ are both continuous functions. Now $x^n(1 - x)$ has a maximum on $[0, 1]$ at $x=\frac{n}{1+n}$ since $\frac{d}{dx}[x^n(1-x)] = -x^n +nx^{n-1}-nx^n = -x-nx+n$ thus $x=\frac{n}{1+n}$.

Then $|x^n(1-x)|<(\frac{n}{n+1})^n(\frac{1}{n+1})<\frac{1}{n+1}<\epsilon.$ Choose $N = \frac{1-\epsilon}{\epsilon}$ therefore for $n>N$ we have $|x^n(1-x)|<\epsilon.$

Why is $\left ( \frac{n}{n+1}\right )^n < 1$?

2. Nov 3, 2013

### vanhees71

I think that's pretty obvious, but one can of course formally prove it:
$$0<1 \; \Rightarrow \; n<n+1.$$
Now since $n+1>0$ you have
$$\frac{n}{n+1}<1.$$
Now multiplying this with $n/(n+1)$ gives
$$\left (\frac{n}{n+1} \right )^2 < \frac{n}{n+1}<1,$$
and in this way you can prove that
$$\left (\frac{n}{n+1} \right )^k<1$$
for all $k \in \mathbb{N}$. Setting $k=n$ gives the inequality you asked for.

3. Nov 3, 2013

### Lee33

vanhees71 - Thanks, that is a nice way to prove it!

Is there another way, for example something similar $\lim_{n\to \infty}(\frac{n}{n+1})^n <1$ implies $\lim_{n\to \infty} \left(1+\frac{1}{n}\right)^n = e > 1$? Or am I wrong?

4. Nov 3, 2013

### dirk_mec1

It has to be valid for all n>N thus your exponential limit is not going to work here.

5. Nov 3, 2013

### Lee33

Dirk-mec1 - So I cant use that? Vanhees71 proof will suffice?

6. Nov 3, 2013

### vela

Staff Emeritus
You have that backwards. The limit
$$\lim_{n\to \infty} \left(1+\frac{1}{n}\right)^n = e$$ implies that
$$\lim_{n\to \infty} \left(\frac{n}{n+1}\right)^n = \frac{1}{e} < 1.$$ You can probably use this fact in an indirect way for your needs. You know that if $n$ is large enough, $\left(\frac{n}{n+1}\right)^n$ will necessarily be less than 1 because the sequence approaches 1/e. With a suitable choice for $N$, you will get the result you need. But why would you want to do that? The proof vanhees provided is much clearer and direct.

7. Nov 3, 2013

### Lee33

Vela - Thanks and you're right, vanhees proof is a lot more clear! Thanks for all the help.