1. ## Series

Consider the serie:
$\displaystyle u(n) := \dfrac{E(10^nx)}{10^n}$
E(x) means the integer part of x (E(5.2) = 5 for example)

The limit of the serie is x.

Now suppose x < x' and show that u(n) < u'(n).

u(n) is the serie of x and u'(n) the serie of x'.

I actually found a way, but it's not really good, because it contains a small contradiction. I wondered whether somebody knows who to get around that contradiction.

This is what I got when I started with the definitions of a serie:

x-x' < u(n) - u'(n) < x-x'

If I ignore the left part it works, but the way it stands there now, it's impossible since a number cannot be strictly smaller than itself.

2. Now suppose x < x' and show that u(n) < u'(n).
Since x<x', we have E(10^n*x)<E(10^n*x'), divide by 10^n and we are done. Or have I misunderstood?

3. God how stupid and I didn't get it!!!!! It's correct. I just have to add that it this is true for n large enough. (If n=1 and you take 5.2 and 5.27 it doesn't work. You have to assume that n > 2 then. for example) But that's it! Thx a 1000 times!!!