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: $\pi/4$.

 

[M98Ranger]

Yes, you can use an integral to show this limit. In fact, this is a common method for evaluating limits of infinite sums. To show this limit, you can use the Riemann sum approach, where you approximate the sum with a definite integral. This is similar to what Dextercioby did in the previous thread.

First, rewrite the sum as:

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

Next, we can use the fact that:

\sqrt{\frac{n^{2}-i^{2}}{n^{2}}} \approx \sqrt{1-\frac{i^{2}}{n^{2}}}

for large n.

Then, we can rewrite the sum as:

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

This is now in the form of a Riemann sum, which can be approximated by a definite integral.

\lim_{n\rightarrow \infty}\left( \frac{1}{n}\cdot\sum_{i=1}^{n}\sqrt{1-\frac{i^{2}}{n^{2}}}\right) \approx \int_{0}^{1}\sqrt{1-x^{2}}dx

This integral can be evaluated using trigonometric substitution, and it will yield the result \frac{\pi}{4}. Therefore, we can conclude that:

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