Infinity

Interesting teaser a friend of mine told me yesterday:

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

As n->infinity what does x tend to?

when n-> to infinity, (1/n) tends to be 0, so (1+0)^n =1

Nope :)

It's e innit?

Yep. Although when I put it into Excel it runs out of accuracy and does some REALLY strange things...check out this graph:

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.

Parksie, I'm sure kedaman didnt mean put your head back in ASCII "Parksie"! I think he meant avatar... jeez, that's scary.

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

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.

As n->infinity x tends to marmalade

*burble*

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

Lt_n->0(1+n)^(1/n) = e


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

Thanks for the tip guv.. :)

Knight