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Quick question on real analysis proof

  1. Nov 3, 2013 #1
    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. jcsd
  3. Nov 3, 2013 #2

    vanhees71

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    I think that's pretty obvious, but one can of course formally prove it:
    [tex]0<1 \; \Rightarrow \; n<n+1.[/tex]
    Now since [itex]n+1>0[/itex] you have
    [tex]\frac{n}{n+1}<1.[/tex]
    Now multiplying this with [itex]n/(n+1)[/itex] gives
    [tex]\left (\frac{n}{n+1} \right )^2 < \frac{n}{n+1}<1,[/tex]
    and in this way you can prove that
    [tex]\left (\frac{n}{n+1} \right )^k<1[/tex]
    for all [itex]k \in \mathbb{N}[/itex]. Setting [itex]k=n[/itex] gives the inequality you asked for.
     
  4. Nov 3, 2013 #3
    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?
     
  5. Nov 3, 2013 #4
    It has to be valid for all n>N thus your exponential limit is not going to work here.
     
  6. Nov 3, 2013 #5
    Dirk-mec1 - So I cant use that? Vanhees71 proof will suffice?
     
  7. Nov 3, 2013 #6

    vela

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    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.
     
  8. Nov 3, 2013 #7
    Vela - Thanks and you're right, vanhees proof is a lot more clear! Thanks for all the help.
     
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