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Thread: The number e as a sequence

  1. #1
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    The number e as a sequence

    I'm not comfortable with a certain proof that the sequence $\displaystyle {y_n} = 1 + \frac{1}{{1!}} + \frac{1}{{2!}} + \cdots + \frac{1}{{n!}}$ converges to the number e. The proof uses the fact that $\displaystyle {y_n}$ is an increasing sequence that is bounded (which you show by a greater geometric series). Therefore it is convergent. Then you expand the the sequence

    $\displaystyle {x_n} = {\left( {1 + \frac{1}{n}} \right)^n} = 1 + 1 + \frac{1}{{2!}}{\left( {1 - \frac{1}{n}} \right)^{}} + \cdots \frac{1}{{n!}}\left( {1 - \frac{1}{n}} \right)\left( {1 - \frac{2}{n}} \right) \cdots \left( {1 - \frac{{n - (n - 1)}}{n}} \right)$

    which we know converges to e. Also, by comparing terms we see that $\displaystyle {x_{n \le }}{y_n}$. So, $\displaystyle {y_n} \ge e$. What remains is to show $\displaystyle {y_n} \le e$ which is the part I don't get. The book says to use the sum of the first $\displaystyle m + 1$ terms in $\displaystyle {x_n}$ where $\displaystyle m < n$.

    So,

    $\displaystyle = 1 + 1 + \frac{1}{{2!}}{\left( {1 - \frac{1}{n}} \right)^{}} + \cdots \frac{1}{{m!}}\left( {1 - \frac{1}{n}} \right)\left( {1 - \frac{2}{n}} \right) \cdots \left( {1 - \frac{{n - (n - 1)}}{n}} \right) < {x_n} < e$

    Now they it says to hold m fixed and let n increase, which gives $\displaystyle {y_m} = 1 + \frac{1}{{1!}} + \cdots \frac{1}{{m!}} \le e$ and therefore $\displaystyle {y_n} \le e$.

    But how is it valid to use a finite # of terms? We have to deal with infinite terms, and I tell myself that y may exceed e in that case.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: The number e as a sequence

    What You can demonstrate without any other 'particulary knowledge' is that the sequence...

    $\displaystyle a_{n}= (1+\frac{1}{n})^{n}$ (1)

    ... converges to a number which is between 2 and 3... and nobody stops You from calling that number $\displaystyle e$...

    As immediate extension You can demonstrate that the sequence...

    $\displaystyle a_{n} (x) = (1+\frac{x}{n})^{n}$ (2)

    ... convergers for any value [real or complex...] of the $\displaystyle x$ and nobody stops You from defining the function...

    $\displaystyle e^{x}= \lim_{n \rightarrow \infty} (1+\frac{x}{n})^{n}$ (3)

    From (3) it is possible to derive all the properties of $\displaystyle e^{x}$, as for example...

    $\displaystyle e^{x}= \lim_{n \rightarrow \infty} \sum_{k=0}^{n} \frac{x^{k}}{k!}$ (4)

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  3. #3
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    Re: The number e as a sequence

    I'm not seeing how this helps me with y<e. I know how to derive (1) from the definition of the derivative but that's it. I haven't covered series yet.
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  4. #4
    MHF Contributor chisigma's Avatar
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    Re: The number e as a sequence

    Quote Originally Posted by zg12 View Post
    I'm not seeing how this helps me with y<e. I know how to derive (1) from the definition of the derivative but that's it. I haven't covered series yet.
    The problem is $\displaystyle how$ to define the number e [which is a number as any other number...] in a precise and [possibly...] simple way... without any knowledge of 'derivatives' but with the only 'four elementary operations' You can demonstrate that the sequence $\displaystyle (1+\frac{1}{n})^{n}$ converges to a number that is between 2 and 3 and You define...

    $\displaystyle e= \lim_{n \rightarrow \infty} (1+\frac{1}{n})^{n}$ (1)

    The same is for...

    $\displaystyle e^{x}= \lim_{ n \rightarrow \infty} (1+\frac{x}{n})^{n}$ (2)

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  5. #5
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    Re: The number e as a sequence

    I'm very dense when it comes to mathematics, so I'm not seeing how it relates to my problem. I can get (2) by L'hopitals rule on it.

    $\displaystyle {\lim }\limits_{x \to \infty } {\left( {1 + \frac{a}{x}} \right)^{bx}} = {e^{ab}}$

    So then how do you go to (4) from there?
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