Finding out the limit: $\lim_{n \to \infty}\frac{1}{n}\sum_{k=0}^{n}\sqrt {k(n-k)}$

Question

Find out the limit of the sum $\lim \frac{1}{n}\sum_{k=0}^{n}\sqrt {k(n-k)}$ as $n$ tends to $\infty.$

Answer 1

Hint. One may write, as $n \to \infty$, $$ \frac{1}{n}\sum_{k=0}^{n}\sqrt {k(n-k)}=\sum_{k=0}^{n}\sqrt {\frac{k}n\left(1-\frac{k}n\right)}=n \cdot u_n $$ then, by recognizing a Riemann sum, as $n \to \infty$, $$ u_n=\frac{1}{n}\sum_{k=0}^{n}\sqrt {\frac{k}n\left(1-\frac{k}n\right)} \to \int_0^1 \sqrt{x(1-x)}\:dx=\frac{\pi}8 $$ yielding, as $n \to \infty$, $$ \frac{1}{n}\sum_{k=0}^{n}\sqrt {k(n-k)}=n \cdot u_n \to \infty. $$

Answer 2

Since $x-x^2$ has a peak at $x=\frac{1}{2}$ and decreases monotonically to $0$ as $x \downarrow 0$ or $x \uparrow 1$, we see $\frac{1}{n}\sum_{k=0}^n \sqrt{k(n-k)} = \sum_{k=1}^{n-1} \sqrt{\frac{k}{n}(1-\frac{k}{n})} \ge \sum_{k=1}^{n-1} \sqrt{\frac{1}{n}(1-\frac{1}{n})} = (n-1)\sqrt{\frac{1}{n}-\frac{1}{n^2}} = \sqrt{\frac{(n-1)^2}{n}-\frac{(n-1)^2}{n^2}} \sim \sqrt{n}$ which diverges