(H.W) What is the limit of $\lim_{n \to \infty} \sum_{k = 1}^{n} (1/\sqrt{(4n^2 - k^2)})$?

Question

Honestly I don't know where to begin with. I though of using tricks like $1/(ab) = (1/a - 1/b) \times1/(b-a)$ but the square root term is bothering me. Any help would be appreciated.

Answer 1

For all $n \in \mathbb{N}^{*}$ $$ \sum_{k=1}^{n}\frac{1}{\sqrt{4n^2-k^2}}=\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\displaystyle \sqrt{4-\left(\frac{k}{n}\right)^2}} $$ Hence

$$ \frac{1-0}{n}\sum_{k=1}^{n}\frac{1}{\displaystyle \sqrt{4-\left(\frac{k}{n}\right)^2}}\underset{n \rightarrow +\infty}{\rightarrow}\int_{0}^{1}\frac{\text{d}x}{\sqrt{4-x^2}}=\frac{\pi}{6} $$

Reminder :

For $f : \left[a,b\right] \rightarrow \mathbb{R}$ with $\left(a,b\right) \in \mathbb{R}^2$ ( $a<b$ ) then $$ \frac{b-a}{n}\sum_{k=1}^{n}f\left(a+k\frac{b-a}{n}\right) \underset{n \rightarrow +\infty}{\rightarrow} \int_{a}^{b}f\left(x\right)\text{d}x $$ To fully understand this equality, notice that you are approaching the area of the function defined between $x=a$ and $x=b$ with rectangulars areas. The more $n$ grows, the more you split $\left[a,b\right]$ into thin rectangulars that better approaches this area as the image http://www.mathly.fr/images/sriemann.png shows.