<s>[INST] Sum of floor function of general term While doing some solution to get a general term for some series I encountered at the end with this.
$S(n,i) = \left(\dfrac{n^2+n-2i}{n} - \lfloor\dfrac{n^2+n-2i}{2n}\rfloor \right) \left(\lfloor\dfrac{n^2+n-2i}{2n}\rfloor +1 \right)$ 
, where $i> \dfrac{n}{2}$. How can I get the general term from this summation by geting rid of the floor function. 
 [/INST] What's your problem with floor function?
However you can't get rid of it, but this may help:
$$S(n,i) = \left(\dfrac{n^2+n-2i}{n} - \lfloor\dfrac{n^2+n-2i}{2n}\rfloor \right) \left(\lfloor\dfrac{n^2+n-2i}{2n}\rfloor +1 \right)$$
$$ = \left(\dfrac{n^2+n-2i}{2n} + \dfrac{n^2+n-2i}{2n} - \lfloor\dfrac{n^2+n-2i}{2n}\rfloor \right) \left(\lfloor\dfrac{n^2+n-2i}{2n}\rfloor +1 \right)$$
$$ = \left(\dfrac{n^2+n-2i}{2n} + \epsilon({n^2+n-2i}{2n})\right) \left(\dfrac{n^2+n-2i}{2n} - \epsilon({n^2+n-2i}{2n}) +1 \right)$$
When the $\epsilon$ function returns the decimal part instead of integer part of given number.
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