# Limit of an infinite sum.

1. Mar 14, 2005

### mprm86

$$\lim_{n\rightarrow \infty}\left( \frac{1}{n}\cdot\sum_{i=1}^{n}\sqrt{1-\frac{i^{2}}{n^{2}}}\right)= \frac{\pi}{4}$$

So, how do I show this. Can I do it with an integral (as Dextercioby did)?
This limit came out when I was triyng to derive the formula of the surface area of a sphere in the way Archimedes did.

2. Mar 14, 2005

### Galileo

It looks very much like a Riemann sum.
If you split the interval [0,1] into n subintervals of width $\Delta x = 1/n$, then the right endpoint of the i-th subinterval is $x_i = i/n$.

So the sum can be written:

$$\sum_{i=1}^{n}\sqrt{1-x_i^2}\Delta x$$
So if we let $f(x)=\sqrt(1-x^2)$ then the limit is:

$$\lim_{n\rightarrow \infty}\sum_{i=1}^{n}f(x_i)\Delta x= \int_0^1\sqrt{1-x^2}dx$$
which is the area of a quarter of the unit disc: $\pi/4$.