# proving the existence of "e"

• Sep 22nd 2010, 12:01 PM
sedam7
proving the existence of "e"
hello,

i don't know how to do proving the existence of e ....

okay, i know that it is by definition :

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

or

$\displaystyle e = \sum_{n=0} ^{\infty} \frac {1}{n!}$

I have no idea how to do it, can some one help me with these please :D

P.S. I'm really really sorry if this shouldn't be posted here :D
• Sep 22nd 2010, 01:23 PM
Traveller
e is usually defined as :

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

Its equivalence to the form $\displaystyle e = \sum_{n=0} ^{\infty} \frac {1}{n!}$ can be proved by many methods including using the generalized binomial theorem or directly the Taylor series.

http://en.wikipedia.org/wiki/Taylor_series
• Sep 22nd 2010, 02:02 PM
sedam7
thanks :D

hm... it should not (i think) be shown existence of e using Taylor series, because this we will not do Taylor series for long time ....

i do't have problem using "e" like it is defined... (lol solved a lot problems using that) but when it comes to prove it's existence i'm think i really don't get it... probably i'm having to much problem of figure it out theoretically, and understanding it to :D
• Sep 22nd 2010, 02:18 PM
Plato
I will not give you details. Here is an outline.

I is easy to show that $\dfrac{1}{n+1}<\ln\left(1+\frac{1}{n}\right)<\dfra c{1}{n}$ . (use Napier's inequality)

From that we see $e^{\frac{1}{n+1}}<\left(1+\frac{1}{n}\right)

Raise to $n^{th}$ power $e^{\frac{n}{n+1}}<\left(1+\frac{1}{n}\right)^n

Use squeeze play.
• Sep 22nd 2010, 02:29 PM
sedam7
Quote:

Originally Posted by Plato
I will not give you details. Here is an outline.

I is easy to show that $\dfrac{1}{n+1}<\ln\left(1+\frac{1}{n}\right)<\dfra c{1}{n}$ . (use Napier's inequality)

From that we see $e^{\frac{1}{n+1}}<\left(1+\frac{1}{n}\right)

Raise to $n^{th}$ power $e^{\frac{n}{n+1}}<\left(1+\frac{1}{n}\right)^n

Use squeeze play.

thank you very much (Nod)
• Sep 22nd 2010, 09:08 PM
chisigma
If we accept the definition...

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

... what You have to demonstrated it that the limit exists. From the Newton's expansion...

$\displaystyle a_{n}= (1+\frac{1}{n})^{n}= 1 + \frac{\binom{n}{1}}{n} + \frac{\binom{n}{2}}{n^{2}} + \frac{\binom{n}{3}}{n^{3}} + \dots + \frac{\binom{n}{n}}{n^{n}} =$

$\displaystyle = 1 + 1 + \frac{(1-\frac{1}{n})}{2!} + \frac{(1-\frac{1}{n})\ (1-\frac{2}{n})}{3!} + \dots + \frac{(1-\frac{1}{n})\ (1-\frac{2}{n}) \dots (1-\frac{n-1}{n})}{n!}$ (2)

... it is evident that...

a) the sequence $a_{n}$ is increasing with n and for $n>1$ is $a_{n} > 2$...

b) $\displaystyle a_{n} < 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \dots + \frac{1}{n!}$ (3)

Now is $3!= 3\ 2 > 2^{2}, 4!=4\ 3! > 2^{3} , \dots , n!= n\ (n-1)!> 2^{n-1}$ , so that...

$\displaystyle a_{n} < 1 + 1 + \frac{1}{2} + \frac{1}{2^{2}} + \dots + \frac{1}{2^{n-1}} = 1 + \frac{1-\frac{1}{2^{n}}}{1-\frac{1}{2}} = 1 + 2\ (1-\frac{1}{2^{n}}})$ (4)

... and from (4) You derive that for all n is $a_{n}< 3$. The conclusion is that the limit (1) exists and is $2< e < 3$. For the 'pratical' computation of e the (1) is very slow and is better the relation found by Isaac Netwon himself...

$\displaystyle e= \sum_{n=0}^{\infty} \frac{1}{n!}$ (5)

Kind regards

$\chi$ $\sigma$
• Sep 24th 2010, 05:51 AM
HallsofIvy
If you prefer to use $e= \displaystyle\sum_{n=0}^\infty \frac{1}{n!}$ as the definition of e, then proving that e exists is just proving that the series converges- and that is easy to do using the "ratio test".