# A simple way to find $\lim_{n\rightarrow\infty}{\frac{1}{n^2}\sum_{k=1}^n{\sqrt{n^2-k^2}}}$

I was reading an exam paper used to identify gifted high-school students, and I encountered the following problem: $$\lim_{n\rightarrow\infty}{\frac{1}{n^2}\sum_{k=1}^n{\sqrt{n^2-k^2}}}$$ Using standard calculus techniques, this sum can be easily calculated by treating the sum as a riemann sum, and calculating the corresponding integral. Indeed, if we denote $f(x)=\sqrt{1-x^2}$, we get: $$\lim_{n\rightarrow\infty}{\frac{1}{n^2}\sum_{k=1}^n{\sqrt{n^2-k^2}}} = \lim_{n\rightarrow\infty}{\frac{1}{n}\sum_{k=1}^n{f\left(\frac{k}{n}\right)}} = \int_0^1{f(x)dx}=\frac{\pi}{4}$$ Where the integral is calculated by noting that the area under the graph of $f$, between $0$ and $1$, is a quarter of the unit disk.

However, in my country, riemann sums are not taught in high-schools, so using this technique is not an option. Is there a simpler method to calculate this limit?

• If a student is gifted in math, he/she probably self-learned the Riemann sum way before entering high school... – achille hui May 28 '15 at 11:53

You can avoid talking about integrals if you can talk about areas. Then the sum is a sum of the areas of small rectangles that approximate a quarter of the unit disk. That is the real leap. Perhaps this is enough.

You can still write

$$\frac{1}{n} \sum_{k=1}^n\sqrt{1-\frac{ k^2}{n^2 } }$$

This is just the "average" height of a semicircle. The diameter is 2 and the area is $\frac{\pi }{2 }$.

Then using $A=b \times h$ the average must be $\frac{\pi}{4}$.

The tools to really formalize these dont come until 2nd or 3rd year university. Depending on where you study.

Uh you can estimate the error between the semicircle and the rectangle approximation using very small triangles. It will be quite small and show their total area tends to 0.

The base of the triangle is $\frac{1}{n}$. The height is

$$\left|\sqrt{1-\frac{ k^2}{n^2 }}- \sqrt{1-\frac{ (k+1)^2}{n^2 }}\right| < \frac{2}{\sqrt{n}}$$

So the total error is smaller than $\frac{1}{\sqrt{n}}$