**
___________________________________
| |
| oo |
| --- |
| \ 1 |
| | --- = lim ( 1 + 1/n )**^{n} |
| / n! n->oo |
| --- |
| n=0 |
|___________________________________|

I did a

proof (demonstration, actually, convergance isn't formally considered) of this while paying quite much too little

attention in

grade 11 math... I don't know whose it actually is.

Probably no-one's, it's really

straightforward. Phun, It's done from

a high school perspective, so excuse the

conceptual in

elegance.

Now, since

n is

approaching infinity, applying the

Binomial Theorem to the

right side of the above

equation looks promising... It'll give you a

sum of an

infinite number of

terms. Let's take a look, see.

**The Binomial Theorem:**
n
---
\ n!
(a+b)^{n} = | ----------- * a^{n-k} * b^{k}
/ k!*(n-k)!
---
k=0

**Let's calculate a few terms of the expansion of e, using the theorem, and starting at t**_{n} (k = n).
**t**_{n} (k = n)
___________________________________________________
| |
| n! |
| (1/n+1)^{n} = lim ----------- * (1/n)^{n-n} * (1)^{n} |
| n->oo n!*(n-n)! |
|___________________________________________________|

**Clean it up, and it all comes out to 1 (1/(0!)).** **t**_{n-1} (k = n-1)
_____________________________________________________________
| |
| n! |
| (1/n+1)^{n} = lim ------------------- * (1/n)^{n-(n-1)} * (1)^{n-1} |
| n->oo (n-1)!*(n-(n-1))! |
|_____________________________________________________________|
n(n-1)!
(1/n+1)^{n} = lim ------------- * (1/n)^{1} * (1)^{n-1}
n->oo (n-1)!*1!
(1/n+1)^{n} = lim n * (1/n) * 1 = 1
n->oo

**...and 1 is 1/(1!)**
**t**_{n-2} (k = n-2)
______________________________________________________________
| |
| n! |
| (1/n+1)^{n} = lim ------------------- * (1/n)^{n-(n-2)} * (1)^{n-2} |
| n->oo (n-2)!*(n-(n-2))! |
|______________________________________________________________|
n(n-1)(n-2)!
(1/n+1)^{n} = lim -------------- * (1/n)^{2} * (1)^{n-2}
n->oo (n-2)!*2!
n(n-1) 1
(1/n+1)^{n} = lim -------- * --- * 1
n->oo 2! n^{2}

We can use the limit product law to rearrange this and divide it into two seperate limits:

1 n(n-1)
(1/n+1)^{n} = lim --- * lim -------- * 1
n->oo 2! n->oo n^{2}

The first limit is simply 1/2!, by the constant limit theorem. The second can easily be show to be 1 by multiplying through by 1/n and taking the limit... intuitively, think of how the ratio between n and n-1 shrinks to zero as n approaches infinity.

A similar argument can be used for all following terms

oo
---
1 1 1 1 \ 1
lim ( 1 + 1/n )^{n} = --- + --- + --- + --- + ... = | ---
n->oo 0! 1! 2! 3! / n!
---
n=0

**Whee.**