If we have a sequence of 1's, why is the partial sum of it equal to a floor function? If $a_n = 1$
then why is $$A_n = \sum_1^n a_n = 1+ 1+ ... + 1$$
equal to the floor function $x + O(1)$?  
I got this from Wikipedia's page regarding the Abel summation formula.
Thanks,
 A: Based on your description, what you are really asking is different from the formula
$$A_n = \sum_1^n a_n = 1+ 1+\dots + 1$$
which you in the original version of your post.
You wrote that you have the formula from Wikipedia article on Abel's summation formula (here is link to the current revision at the time of this post.) 
We can see there the formulas 
$$A(x)=\sum_{1\le n\le x} a_n.$$
The article also contains the claim that for $a_n=1$ we get $$A(x)=\lfloor x \rfloor.$$
So I assume this is what you are asking about.
The argument why this works is pretty simple. The sum
$$A(x)=\sum_{1\le n\le x} 1$$
simply counts the number of integers $n$ such that $1\le n \le x$. (Since every such integer adds $+1$ to the sum.)
And the floor function is precisely the number of such integers. (Floor function is equal to the integer $k$ such that $k\le x < k+1$. Which means that for every positive integer $n\le k$ we have also $n\le x$. But for any integer $n>k$ we have $n>x$. So we are simply counting positive integers which are less than or equal to $k$. The number of such integers is $k=\lfloor x\rfloor$.) 
