How to compute $\lim\limits_{n\to\infty}\frac{1}{\sqrt{4n^2-1^2}}+\dots+\frac{1}{\sqrt{4n^2-n^2}}$.

I want to compute $$\lim_{n\to\infty}\sum_{k=1}^n \frac1{\sqrt{4n^2-k^2}}.$$

I found it on this question and the exercise appears to be from previous years of a Latvian competition.

I tried writing $\frac1{\sqrt{4n^2-k^2}}=((2n-k)(2n+k))^\frac{-1}2$ and wanted to continue with partial fractions but the square root is annoying. Also, I thought about $$\frac1{\sqrt{4n^2-k^2}}=\frac{\sqrt{4n^2+k^2}}{\sqrt{16n^4-k^4}}$$ but I don't think that this can help me.

Using Riemann sums:

$$\lim_{n\to\infty}\sum_{k=1}^n \frac1{\sqrt{4n^2-k^2}} =\lim_{n\to\infty}\frac1n\sum_{k=1}^n \frac1{\sqrt{4-\left(\frac kn\right)^2}} =\int_0^1 \frac1{\sqrt{4-x^2}} =\arcsin\left(\frac x2\right)\bigg\vert_0^1 =\frac\pi6. $$

$$\lim_{n\to\infty}\sum_{k=1}^n \frac1{\sqrt{4n^2-k^2}}=\lim_{n\to\infty}\sum_{k=1}^n \frac1{\sqrt{4-\left(\frac{k}{n}\right)^2}}\frac{1}{n}=\int\limits_0^1\frac{1}{\sqrt{4-x^2}}dx.$$ Can you end it now?

Since$$\sum_{k=1}^n\frac1{4n^2-k^2}=\sum_{k=1}^n\frac1n\times\frac1{\sqrt{4-\left(\frac kn\right)^2}},$$this is a Riemann sum. The limit is $\displaystyle\int_0^1\frac{\mathrm dx}{\sqrt{4-x^2}}=\frac\pi6$.

You can rewrite it to become $$\lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^n \frac{n}{\sqrt{4n^2-k^2}} = \lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^n \frac{1}{\sqrt{4-\left(\frac{k}{n}\right)^2}}$$

This is the Riemann sum for $\frac{1}{\sqrt{4-x^2}}$ from $0$ to $1$. Therefore, the limit of the sum is $$\int_0^1 \frac{1}{\sqrt{4-x^2}} dx$$

Integrating, we get $$\arcsin \left( \frac{x}{2} \right) \bigg\vert_0^1 = \arcsin\left(\frac{1}{2}\right) - \arcsin\left(\frac{0}{2} \right) = \frac{\pi}{6}$$