Fuente TeX:

f(n+\frac{1}{2}) = \frac{(n+\frac{1}{2})^2 - \lfloor n + \frac{1}{2} \rfloor^{2}}{2(n+\frac{1}{2})} = \frac{n^2 + n +\frac{1}{4}-n^2}{2(n+\frac{1}{2})}=\frac{n+\frac{1}{4}}{2n + 1} \geq \frac{n}{3n} = \frac{1}{3}