___________________________________
| |
| 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 = | ----------- * an-k * bk
/ k!*(n-k)!
---
k=0
Let's calculate a few terms of the expansion of e, using the theorem, and starting at tn (k = n).
tn (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!)). tn-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!)
tn-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! n2
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 n2
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.