Limit of an infinite sum.

[mprm86]

Sorry, i had already post a thread about this, but it was worng. As Dextercioby said,

\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.

[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.