formula of sum the nimbers

If have well-known formula $(n + 1) n / 2 = 1 + 2 +\cdots+ n$. If the difference between the closest numbers smaller, will obtain, for example $(n + 0,1) n / (2 \cdot 0,1) = 0,1 + 0,2 +\cdots+ n$. Now if the difference between the closest numbers the smallest possible, will obtain $(n + 0,0\ldots1) n / (2 \cdot 0,0\ldots 1) = 0,0\ldots1 + 0,0\ldots2 + \ldots + n$, so can conclude $n ^ 2 / 2 = (0,0\ldots1 + 0,0\ldots2 + \cdots + n) / 0,0\ldots1$ whether conclude is correct?

EDITED VERSION:

If have well-known formula $\frac{(n + 1)n}2 = 1 + 2 +\dots+ n$.

If the difference between the closest numbers smaller, will obtain, for example $\frac{(n + 0,1) n}{2.0,1} = 0,1 + 0,2 +\dots + n$.

Now if the difference between the closest numbers the smallest possible, will obtain $\frac{(n + 0,0\dots1) n}{2 . 0,0\dots 1} = 0,0\dots 1 + 0,0\dots 2 + \dots + n$ , so can conclude $\frac{n ^ 2}2 = \frac{0,0\dots1 + 0,0\dots2 + \dots + n}{0,0\dots1}$ whether conclude is correct?

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By comma symbol, you mean decimal point? –  user18325 Dec 17 '11 at 7:40
@ZeeshanMahmud Comma is used as decimal separator in many countries. –  Martin Sleziak Dec 17 '11 at 7:48
Marko: I guess both these account belong to you: math.stackexchange.com/users/20189/marko and math.stackexchange.com/users/21380/marko; Maybe it would be good for you to register, so that you can better follow all questions you posted. (After you do it, you can even ask moderators to merge you account with the older unregistered ones.) –  Martin Sleziak Dec 17 '11 at 7:52
shiver me nimbers –  jspecter Dec 17 '11 at 8:00
Marko: I've tried to edit your question using TeX for better readability. (And maybe some other users will improve it a little more.) You should check whether I did not change the meaning of your question, unintentionally. If you're satisfied with the edited version, you can remove the original one. –  Martin Sleziak Dec 17 '11 at 8:07

If we denote $k=n/\alpha$ then we get $$\sum\limits_{i=1}^k i\alpha=\frac{k(k+1)}2 \alpha = \frac{k\alpha(k\alpha+\alpha)}{2\alpha} = \frac{n(n+\alpha)}{2\alpha}.$$ For $\alpha$ of the form $10^{-s}$, i.e. $0,00\dots01$, this is precisely the first part of your question.
I am not sure I understand what you mean by the last part of your question, but if you want to say that $\sum\limits_{i=1}^k i\alpha=\frac{k(k+1)}2$ is approximately $\frac{n^2}{2\alpha}$, in some sense it is true, since the error is: $$\frac{n(n+\alpha)}{2\alpha} - \frac{n^2}{2\alpha} = \frac {n\alpha}{2\alpha}=\frac n2,$$ so the error has smaller order than the sum. (The error is $o(n^2)$, if you're familiar with this notation.)
If you know something about Riemann integral, you may notice that the sum $$\alpha \sum\limits_{i=1}^k i\alpha$$ is in fact upper Riemann sum for the function $f(x)=x$ on the interval $[0,n]$ (and for uniform partition of this interval). To see this just notice that the lengths of the intervals of partitions is $\alpha$ and the value at the end of the interval $[(i-1)\alpha,i\alpha]$ is $i\alpha$.
So it is expected that this sum is approximately $\frac{n^2}2$ for small values of $\alpha$. Your sum is $\alpha$-times smaller.