If x ∈ R, we write q = [x], the integer part of x such that x−1 < q ≤ x.
Thus [x] ≤ x. Define $x_n = n^2/(2n+1) -[n/2]$,where [n/2] is the integer part of n/2.
I should calculate inf $x_n$ and sup $x_n$.
I traced the sequence [1/3,-1/5,2/7,-2/9,3/11...]
The supremum $x_n$ is 1/3, but what about the infimum?
When $n$ is odd, $\lfloor n/2\rfloor=(n-1)/2$ , so $x_n=\dfrac{n^2}{2n+1}-\dfrac{n-1}2=\dfrac{n+1}{4n+2}>0$.
when $n$ is even, $\lfloor n/2\rfloor=n/2$, so $x_n=\dfrac{n^2}{2n+1}-\dfrac n2=-\dfrac n{4n+2}$.
Can you now show $\inf x_n=-\dfrac14$?