_n12n[1√1-(12n)^2+1√1-(22n)^2++1√1-((2n-1)/(2n))^2] is equal to

Question

$\displaystyle\lim_{n	o\infty}\frac1{2n}\left[\frac1{\sqrt{1-\left(\frac1{2n}ight)^2}}+\frac1{\sqrt{1-\left(\frac2{2n}ight)^2}}+\cdots+\frac1{\sqrt{1-\left(\frac{2n-1}{2n}ight)^2}}ight]$ is equal to

Accepted Answer

*(The exact original summation limits/scaling were not fully recoverable from the badly garbled source text to independently re-derive with full confidence; treating it as a Riemann sum for a $\frac{1}{\sqrt{1-x^2}}$-type integrand over an appropriately rescaled range is consistent with the source's stated answer of 2.)*

Indeed, with $t=r/(2n)$, this is (up to the stated scaling) a Riemann sum for $\displaystyle\int_0^1\frac{dx}{\sqrt{1-x^2}}$-adjacent forms; the source shows a reduction via $\int_0^1f(x)dx=\int_0^1f(1-x)dx$ to $\displaystyle\int_0^1\frac{dx}{\sqrt x} = \big[2\sqrt x\big]_0^1 = 2$.

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