1. Mar 12, 2005

### mprm86

limit when n goes to infinity of the sum from i=1 to n of 1/n * sqrt(1-i^2/n^2) equals to zero.

Sorry, i havent learned yet to use that Tex thing.

2. Mar 12, 2005

### physicsCU

Not sure of the formal way, but here is what I get.

The inside of the sqrt goes to zero because n^2 goes to infinity, making the interior go to zero. 0/n as n -> infinity is still 0.

3. Mar 12, 2005

### dextercioby

Yes,but it's an infinite sume of "zero-s"...Are us sure you're referring to

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

My maple says it's $$\frac{\pi}{4}$$...

Daniel.

4. Mar 12, 2005

### dextercioby

It is.I'm a genius.That limit (together with the sum) can be put in connection to the Riemann sum of the integral (it actually is)

$$I=:\int_{0}^{1}\sqrt{1-x^{2}} \ dx =\frac{\pi}{4}$$...

You must be talking about something else...

Daniel.

5. Mar 13, 2005

### physicsCU

I think the 1/n is inside the summation, but I am not totally sure. Can we get clarity from the thread starter.

6. Mar 13, 2005

### Data

You can take the $$\frac{1}{n}$$ out of the summation because it doesn't depend on $$i$$. As stated above by Daniel,

$$\lim_{n \rightarrow \infty} \left(\sum_{i=1}^n \left[\frac{1}{n} \sqrt{1 - \frac{i^2}{n^2}}\right]\right) = \int_0^1 \sqrt{1-x^2} \ dx = \frac{\pi}{4}$$

Here $$\frac{1}{n}$$ is the width $$\Delta x$$ of the intervals in the Riemann sum, and $$\frac{i}{n}$$ is the $$x_i^*$$ (and you will note that it is indeed always in the appropriate interval).

Last edited: Mar 13, 2005
7. Mar 13, 2005