Limited Series

Calculus Level 4

$\large \lim_{n\to \infty}\sum_{k=1}^{n}\frac{\sqrt{n^2-k^2}}{n^2}=?$

\pi

\frac{\pi}{4}

\frac{\pi}{2}

None of These Does Not Exist

1 solution

Rishabh Jain
Mar 29, 2017

$\lim_{n\to \infty}\frac 1n\sum_{k=1}^{n}\sqrt{1-\left(\frac{k}{n}\right)^2}$

Using Reimann Sums , this is:

$\int_0^1\sqrt{1-x^2}\mathrm{d}x$

Substitute $x=\sin \theta$ so that integration is:

$\int_0^{\frac{\pi}2}\cos \theta\times \cos \theta \mathrm{d}\theta$

$=\int_0^{\frac{\pi}2}\dfrac{1+\cos 2\theta}2 \mathrm d\theta$

$=\left[\dfrac{\theta+\frac{\sin 2\theta}2}2\right]_0^{\frac{\pi}2}$ $=\boxed{\dfrac{\pi}4}$

Nice solution. A shortcut would be to note that $\displaystyle\int_{0}^{1} \sqrt{1 - x^{2}} dx$ is the area of a quarter of a unit circle.

Brian Charlesworth - 4 years, 2 months ago

Precisely what I did. :-)

Akeel Howell - 4 years, 1 month ago

Right... Thanks

Rishabh Jain - 4 years, 2 months ago

Good shortcut

Aman Bhandare - 4 years, 1 month ago