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parksie
Mar 17th, 2001, 08:50 AM
Interesting teaser a friend of mine told me yesterday:

x = (1 + (1/n))^n

As n->infinity what does x tend to?

klorel82
Mar 17th, 2001, 10:01 AM
when n-> to infinity, (1/n) tends to be 0, so (1+0)^n =1

parksie
Mar 17th, 2001, 10:06 AM
Nope :)

HarryW
Mar 17th, 2001, 11:05 AM
It's e innit?

parksie
Mar 17th, 2001, 02:04 PM
Yep. Although when I put it into Excel it runs out of accuracy and does some REALLY strange things...check out this graph:

Guv
Mar 18th, 2001, 11:43 PM
I suspect that there is an error in your Excel algorithm.

When I use MathCad7 or my HP calculator, the series seems to converge to e. I am sure that Excel could do likewise if given a correct formula or algorithm.

Sastraxi
Mar 19th, 2001, 06:31 PM
Parksie, I'm sure kedaman didnt mean put your head back in ASCII "Parksie"! I think he meant avatar... jeez, that's scary.

Mar 22nd, 2001, 12:13 AM
x = (1 + (1/n))^n

As n->infinity what does x tend to?

Technically, if n is greater then infinity, then adding 1 to the infinit # would excede it's boundry to the infinit power.

and x would be the end result.

Am I right or wrong here?
Knight Vision

Guv
Mar 22nd, 2001, 12:50 AM
Knight_Vision: Limit[ (1 + 1 / n)^n ] = e as n grows without bound.

e = 2.718281828. . . where 1828 does not repeat again. It is not difficult to show that the limit is the sum of the following infinite series.

1 + 1/1 + 1/2 + 1/6 + 1/24 + 1/120 + 1/720 . . . +1/n! . . .

The denominators are factorials: 5! = 5*4*3*2*1

Please note that I did not mention infinity, choosing to say grows without bound. Set theory deals with infinity, other branches of mathematics avoid it.

Talking about infinity without all sorts of background axioms & definitions almost always leads to nonsense. Unless you study Cantor's work on transfinite numbers, you should avoid talking about infinity. Stick with terms like grows without bound and let it go at that.

Mar 22nd, 2001, 04:56 AM
As n->infinity x tends to marmalade

*burble*

Active
Mar 22nd, 2001, 07:20 AM
x = (1 + (1/n))^n

As x -> infinity e for all values of n whether it is +ve -ve
or in a rational form.

This is similar to..writing

Ltn->0(1+n)(1/n) = e


This is what I remember when I studied in Pre-Calculus.

Mar 22nd, 2001, 07:24 AM
Thanks for the tip guv.. :)

Knight