Results 1 to 12 of 12

Math Help - Sequence convergence proof

  1. #1
    Junior Member
    Joined
    Apr 2010
    Posts
    36

    Sequence convergence proof

    What is the simplest way to proove

    \lim_{n \to \infty} (1 + \frac{1}{n})^n = e

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,564
    Thanks
    1425
    This is the value of e by definition.

    Recall that e is defined to be the base of the exponential function such that this function is its own derivative.

    So if \frac{d}{dx}(e^x) = e^x then we have

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

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

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

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

    Clearly for this equation to hold true, that must mean

    \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 = \lim_{h \to 0}h

    \lim_{h \to 0}e^h = \lim_{h \to 0}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}}

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

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


    Now supposing we let n = \frac{1}{h}, clearly as h \to 0, n \to \infty, so this means

    e = \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n.


    Q.E.D.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
    Joined
    Dec 2009
    From
    Russia
    Posts
    1,506
    Thanks
    1
    Quote Originally Posted by losm1 View Post
    What is the simplest way to proove

    \lim_{n \to \infty} (1 + \frac{1}{n})^n = e

    Thanks
    Hello!

    First of all e is just a sing for "special" number.

    So, what you have is a infinite sequence of numbers.

    Theorem:
    If infinite sequence of numbers is bounded and monotonic, the our sequence is converges, in other words there is some number which is called the limit of infinite sequence of numbers.

    From our theorem we need to prove two things:

    1. Boundedness
    2. Monotonicity.(?)

    Prove: using Inequality of arithmetic and geometric means - Wikipedia, the free encyclopedia

    1. \sqrt[n+1]{(1+\frac{1}{n})^n\cdot 2} < \frac{n+1+2}{n+1}<3

    2. \sqrt[n+1]{(1+\frac{1}{n})^n\cdot 1}< \frac{1+n+1}{n+1}

    (1 + \frac{1}{n})^n<(1 + \frac{1}{n+1})^{n+1}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    The correct setting of the problem is the proof that the sequence a_{n} = (1+\frac{1}{n})^{n} converges to a finite number that we call 'e' and is 2 < e < 3. All other results, like for example the well known relation \frac{d}{dx} e^{x} = e^{x} used in the 'demostration', are derived from this fundamental preliminary result and not vice-versa...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,564
    Thanks
    1425
    Incorrect, e is defined to be the base of the exponential function such that this function is its own derivative.

    Therefore the setting of \frac{d}{dx}(e^x) = e^x from the definition and finding an expression for e from the definition is perfectly acceptable.

    But yes, you are correct in saying that when you get to the expression you need to prove \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n exists and can be evaluated.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    Quote Originally Posted by Prove It View Post
    Incorrect, e is defined to be the base of the exponential function such that this function is its own derivative... therefore the setting of \frac{d}{dx}(e^x) = e^x from the definition and finding an expression for e from the definition is perfectly acceptable...
    Very well! ... in order to sustain Your thesis You have to demonstrate that is \frac{d}{dx} e^{x} = e^{x} independently by the fact that \lim_{r \rightarrow \infty} (1+\frac{1}{r})^{r} exists finite and it is between 2 and 3...

    You are welcome!...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by Prove It View Post
    Incorrect, e is defined to be the base of the exponential function such that this function is its own derivative.

    Therefore the setting of \frac{d}{dx}(e^x) = e^x from the definition and finding an expression for e from the definition is perfectly acceptable.

    But yes, you are correct in saying that when you get to the expression you need to prove \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n exists and can be evaluated.
    When replying to a post please quote the post you are responding to, it avoids confusion.

    CB
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by chisigma View Post
    Very well! ... in order to sustain Your thesis You have to demonstrate that is \frac{d}{dx} e^{x} = e^{x} independently by the fact that \lim_{r \rightarrow \infty} (1+\frac{1}{r})^{r} exists finite and it is between 2 and 3...

    You are welcome!...

    Kind regards

    \chi \sigma
    There are a number of ways of defining "e", since we regard what was to be proven as a definition it is obvious that this is to be proven from some other definition. Two that come to mind immediately are the base of natural logarithms and the solution of the IVP (differential equation).

    Earlier posts in this thread have done what is required admirably.

    Your post quoted above comes close to being insulting and will only lead to trouble.

    CB
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,564
    Thanks
    1425
    Very well...

    Assume that you can write e^x as a polynomial c_0 + c_1x + c_2x^2 + c_3x^3 + \dots.

    Therefore

    e^x = c_0 + c_1x + c_2x^2 + c_3x^3 + \dots

    By setting x = 0 we can see c_0 = 1.

    We also know that \frac{d}{dx}(e^x) = e^x by definition, so differentiating both sides gives...

    e^x = c_1 + 2c_2x + 3c_3x^2 + 4c_4x^3 + \dots.

    Let x = 0 and we can see c_1 = 1.


    Differentiate both sides:

    e^x = 2c_2 + 2\cdot 3c_3x + 3\cdot 4c_4x^2 + 4\cdot 5 c_5x^3 + \dots.

    Let x = 0 and we can see c_2 = \frac{1}{2}.


    Differentiate both sides

    e^x = 2\cdot 3c_3 + 2\cdot 3 \cdot 4 c_4x + 3\cdot 4 \cdot 5c_5 x^2 + 4 \cdot 5 \cdot 6c_6x^3 + \dots.

    Let x = 0 and we can see c_3 = \frac{1}{2\cdot 3}.


    It can be seen that if you were to continue this way you would get

    e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{3\cdot 2} + \frac{x^4}{4\cdot 3\cdot 2} + \frac{x^5}{5\cdot 4 \cdot 3\cdot 2} + \dots

    e^x = \sum_{r = 0}^{\infty}\frac{x^r}{r!}.


    By letting x = 1 we find

    e^1 = \sum_{r = 0}^{\infty}\frac{1}{r!}.


    But we have also shown that e = \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n.

    Therefore \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n = \sum_{r = 0}^{\infty}\frac{1}{r!}, proving that this limit exists, and by evaluating the sum to an acceptable number of terms, can be seen to be 2.718\,281\,828\,459\,045\,235\,36\dots.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    Quote Originally Posted by Prove It View Post

    ... we also know that \frac{d}{dx}(e^x) = e^x by definition...
    ... only a minor detail [among many other details...] : You were asked to prove that is \frac{d}{dx} e^{x} = e^{x} , so that Your 'answer' that it is true 'by definition' doesn't seem to be fully adequate...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Senior Member eumyang's Avatar
    Joined
    Jan 2010
    Posts
    278
    Thanks
    1
    Quote Originally Posted by chisigma View Post
    ... only a minor detail [among many other details...] : You were asked to prove that is \frac{d}{dx} e^{x} = e^{x} , so that Your 'answer' that it is true 'by definition' doesn't seem to be fully adequate...
    Maybe I'm dense, but I thought the original question was to prove
    \lim_{n \to \infty} (1 + \frac{1}{n})^n = e.

    The original poster wanted the simplest way to prove this, and he/she did not say anwhere that we could not use the \frac{d}{dx} e^{x} = e^{x} definition. So I don't know why the insistence to prove it without this definition.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,564
    Thanks
    1425
    Quote Originally Posted by chisigma View Post
    ... only a minor detail [among many other details...] : You were asked to prove that is \frac{d}{dx} e^{x} = e^{x} , so that Your 'answer' that it is true 'by definition' doesn't seem to be fully adequate...

    Kind regards

    \chi \sigma
    Incorrect once again. The question was asking to prove \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n = e.

    Using the definition \frac{d}{dx}(e^x) = e^x is once again appropriate.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proof of Convergence for Recursive Sequence
    Posted in the Calculus Forum
    Replies: 9
    Last Post: June 4th 2010, 01:59 AM
  2. Convergence sequence proof
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: May 16th 2010, 03:05 AM
  3. Proof convergence sequence of sets
    Posted in the Discrete Math Forum
    Replies: 8
    Last Post: February 26th 2010, 09:06 AM
  4. Replies: 4
    Last Post: April 28th 2009, 07:29 PM
  5. Sequence Convergence proof maybe?
    Posted in the Calculus Forum
    Replies: 9
    Last Post: October 30th 2008, 12:39 PM

Search Tags


/mathhelpforum @mathhelpforum