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.

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 $\displaystyle\int_0^1\frac{\mathrm dx}{\sqrt{4-x^2}}=\frac\pi6$.
