Limit of Sum as n Approaches Infinity: 1/n * sqrt(1-i^2/n^2) = 0

[mprm86]

Please show that:

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 haven't learned yet to use that Tex thing.

 

[physicsCU]

Please show that:

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 haven't learned yet to use that Tex thing.

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.

 

[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.

 

[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.

 

[physicsCU]

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.

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

 

[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).

 

[physicsCU]

Ah, my bad.

I don't really deal with limits like that, so I was basing what I said on what I know.

I was wrong though.