Quick question on real analysis proof

[Lee33]

Homework Statement

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

 

[vanhees71]

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

 

[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?

 

[dirk_mec1]

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

 

[Lee33]

Dirk-mec1 - So I can't use that? Vanhees71 proof will suffice?

 

[vela]

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?

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.

 

[Lee33]

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