The number e as a sequence

**zg12** · June 28th 2011, 06:35 AM

I'm not comfortable with a certain proof that the sequence ${y_n} = 1 + \frac{1}{{1!}} + \frac{1}{{2!}} + \cdots + \frac{1}{{n!}}$ converges to the number e. The proof uses the fact that ${y_n}$ 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

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

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

So,

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

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

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.

**chisigma** · June 28th 2011, 07:15 AM

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

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

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

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

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

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

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

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

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

Kind regards

$\chi$ $\sigma$

**zg12** · June 28th 2011, 08:01 AM

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.

**chisigma** · June 28th 2011, 08:15 AM

Originally Posted by zg12

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 $how$ 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 $(1+\frac{1}{n})^{n}$ converges to a number that is between 2 and 3 and You define...

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

The same is for...

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

Kind regards

$\chi$ $\sigma$

**zg12** · June 28th 2011, 08:44 AM

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.

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

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

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