$$= \sum_{r=1,3,5 \ldots}^{n-1} \frac{n^2}{({n^2 + r^2})^{\frac32}} = \sum_{r=1,3,5 \ldots}^{n-1} \frac{1}{n({1 + (r/n)^2})^{\frac32}} = \int_{0}^{1}\frac{dx}{({1+x^2})^{\frac32}} = \frac1{\sqrt2}$$ (obtained via subsitution $x=\tan\vartheta$)
Does this answer seems fine?
Minor corrections $$ \sum_{r=1,3,5,\ldots}^{n-1} \frac{n^2}{({n^2 + r^2})^{\frac32}} = \frac{1}{n}\sum_{k=1}^{n/2} \frac{1}{\big({1 + \big(\frac{2k-1}{n}\big)^2}\big)^{\frac32}} \to\frac{1}{2}\int_{0}^{1}\frac{dx}{({1+x^2})^{\frac32}} = \frac1{2\sqrt2} $$