Results 1 to 9 of 9

Math Help - proof of that the definition limit of "e" exists... please help!

  1. #1
    Newbie
    Joined
    Mar 2007
    Posts
    24

    proof of that the definition limit of "e" exists... please help!

    Hi,

    I'm reviewing some basic calculus concepts, and I have the following questions that will probably be very easy for you guys to answer, although I'm stuck since a long time on them.

    Given the usual limit that is used in the definition of "e": lim (n->infinity) (1+1/n), how do we prove that the limit exists and it is finite?

    Can we use the binomial theorem expansion for a finite n, and prove that the coefficients tend to be 1/k! for n that goes to infinity, so we get the Taylor expansion? I'd like to see a rigorous proof of this derivation of e's expression as a power series, we shouldn't forget to prove that the coefficients of the binomial expansion don't diverge (in fact, i guess they should converge) as n goes to infinity...

    Or without expanding it, can we prove the following statements:

    let a(n)=(1+1/n)^n

    1-prove that a(n) is strictly increasing
    2-prove that a(n) is limited above by a finite positive number L (for instance, we could try to prove it for L=3 or L=4): a(n)<L for all n.

    Final questions:

    A-if we manage to prove the two statements above, then did we successfully prove that the limit exists?
    B-is there a better way than this to prove that the limit exists?

    Thanks! And please excuse me for the length of the post...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,646
    Thanks
    1596
    Awards
    1
    Using the notation e^x=\exp(x) we show it.
    It is easy to use Napierís inequality to show that \dfrac{1}{n+1}\le\ln\left(1+\frac{1}{n}\right)\le\  dfrac{1}{n}.

    Raise e to those powers \exp\left(\frac{1}{n+1}\right)\le\left(1+\frac{1}{  n}\right)\le\exp\left(\frac{1}{n}\right)..

    Raise each of those to the nth power
    \exp\left(\frac{n}{n+1}\right)\le\left(1+\frac{1}{  n}\right)^n\le e..

    It is squeezed between e & e.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2007
    Posts
    24
    Thank you for answering so soon, the proof is very simple, but how do you define ln(x) without proving the limit of e first? I mean, wouldn't it be circular logic?

    We should find a way to prove it without using the natural logarithm...

    Alternatively, we could define ln(x) as the integral of 1/t from 1 to x, and then napier's inequality should be simple to infer, but we need to use a integral definition...

    there should be an elementary method to prove it...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,646
    Thanks
    1596
    Awards
    1
    There are so many different ways to do this.

    Notice that for n\ge 2 we have  \[<br />
\left( {1 - \frac{1}{{n^2 }}} \right)^n  \geqslant \left( {1 - \frac{1}<br />
{n}} \right)\; \Rightarrow \;\left( {1 + \frac{1}{n}} \right)^n  \geqslant \left( {1 + \frac{1}{{n - 1}}} \right)^{n - 1} .

    So the sequence  \left( {1 + \frac{1}{n}} \right)^n is increasing and bounded above. That means it converges.

    Call that limit e.

    EDIT: If you need to show it is bounded.
    \left( {1 + \frac{1}<br />
{n}} \right)^n  = \sum\limits_{k = 0}^n {C(n,k)\left( {\frac{1}<br />
{n}} \right)^k }  < 1 + \left( {1 + \frac{1}<br />
{2} +  \cdots  + \frac{1}<br />
{{2^{n - 1} }}} \right) < 3
    Last edited by Plato; October 1st 2010 at 11:27 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2007
    Posts
    24
    Ok thank you very much, I understood how you deduced the facts from those inequalities, I will work to prove them.

    You have been very kind, thanks again.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,535
    Thanks
    778
    Here is a couple of links from PlanetMath.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Mar 2007
    Posts
    24
    Great, thanks again!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,547
    Thanks
    1418
    By definition, e is the base of the exponential function that is its own derivative.

    Therefore \frac{d}{dx}(e^x) = e^x.

    If we wish to evaluate e, we need to use this definition.


    \frac{d}{dx}(e^x) = \lim_{h \to 0}\frac{e^{x + h} - e^x}{h}

     = \lim_{h \to 0}\frac{e^xe^h - e^x}{h}

     = \lim_{h \to 0}\frac{e^x(e^h - 1)}{h}

     = e^x\lim_{h \to 0}\frac{e^h - 1}{h}.


    Since we know by definition that \frac{d}{dx}(e^x) = e^x, that means

    e^x\lim_{h \to 0}\frac{e^h - 1}{h} = e^x

    \lim_{h \to 0}\frac{e^h - 1}{h}= 1

    \lim_{h \to 0}(e^h - 1) = \lim_{h \to 0}h

    \lim_{h \to 0}e^h = \lim_{h \to 0}(1 + h)

    \lim_{h \to 0}(e^h)^{\frac{1}{h}} = \lim_{h \to 0}(1 + h)^{\frac{1}{h}}

    e = \lim_{h \to 0}(1 + h)^{\frac{1}{h}}.


    Now if we let h = \frac{1}{n}, then as h \to 0, n \to \infty.

    Thus e = \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,693
    Thanks
    1466
    Another useful way to define ln(x) is simply:
    ln(x)= \int_1^x \frac{1}{t}dt

    Since \frac{1}{t} is defined for all non-zero t, this defines ln(x) for all positive x. It is clearly continuous and differentiable for all x. By the fundamental theorem of Calculus, its derivative is 1/x. Since that is positive for all positive x, ln(x) is an increasing function and so a one-to-one function.

    It is not difficulty to prove, by substitutions in the integral, that
    1) For any positive x, ln(1/x)= - ln(x).
    2) For any positive x and y, ln(xy)= ln(x)+ ln(y).
    3) For any positive x and y any real number, ln(x^y)= y ln(x).

    For example, for the last one, ln(x^y) is, by this definition, \int_1^{x^y}\frac{1}{t}dt.

    If y\ne 0, define u= t^{1/y} so that t= u^y, du= y u^{y- 1}du so that \frac{1}{t}dt= \frac{1}{u^y}(y u^{y-1}du)= y \frac{1}{u}du. When t= 1, u= 1^{1/y}= 1 and when t= x^y, u= (x^y)^{1/y}= x so the integral becomes
    ln(x^y)= \int_1^x y \frac{1}{u}du= y\int_1^x \frac{1}{u}du= y ln(x)

    (If y= 0 then x^y= x^0= 1 and ln(x^y)= ln(1)= \int_1^1 \frac{1}{t}dt= 0= 0 ln(x)= y ln(x) still.)

    It follows then that ln(x) is a one-to-one function from the positive real numbers onto the real numbers and so has an inverse function, exp(x), from the real numbers to the positive real numbers.

    Finally, if y= exp(x), then x= ln(y). If x is not 0, divide both sides by x to get 1= \frac{1}{x}ln(y)= ln(y^{1/x}). Going back to the exponential form, y^{1/x}= exp(1) so y= (exp(1))^x.
    (If x= 0, then y= 1 because ln(1)= \int_1^1 \frac{1}{t}dt= 0. So we still have exp(x)= exp(0)= 1= (exp(1))^0.
    Last edited by HallsofIvy; October 2nd 2010 at 02:59 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: April 24th 2011, 07:01 AM
  2. Replies: 1
    Last Post: October 25th 2010, 04:45 AM
  3. Replies: 2
    Last Post: August 26th 2010, 01:34 PM
  4. Rigorous Definition of "Inequality" or "Positive" in the Reals
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: July 22nd 2010, 12:23 AM
  5. A "trivial" limit proof
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 9th 2008, 04:16 AM

Search Tags


/mathhelpforum @mathhelpforum