JEE Advanced 2016 Paper 2 Mathematics: Paragraph 2

Write $a_n=\frac1n\sum_{k=1}^{n}f\left(\frac kn\right)$ with $f(x)=\frac1{1+x^{2}}$, since $\frac{n}{n^{2}+k^{2}}=\frac1n\cdot\frac1{1+(k/n)^{2}}$. By the stated expansion, $a_n=\int_0^1\frac{dx}{1+x^{2}}+\frac{f(1)-f(0)}{2n}+o\left(\frac1n\right)=\frac\pi4+\frac{\frac12-1}{2n}+o\left(\frac1n\right)=\frac\pi4-\frac1{4n}+o\left(\frac1n\right)$. Hence $n\left(\frac\pi4-a_n\right)\to\frac14$. A numerical check at $n=10^{5}$ gives $0.250000\ldots$. Option (A) results from using $\frac{f(1)-f(0)}{n}$ with the wrong factor, option (B) from halving again, and option (D) from using the limit $\frac\pi4$ itself.

Here $c_n=\frac1n\sum_{k=1}^{n}g\left(\frac kn\right)$ with $g(x)=\frac{x}{1+x^{2}}$, because $\frac{k}{n^{2}+k^{2}}=\frac1n\cdot\frac{k/n}{1+(k/n)^{2}}$. As $n\to\infty$ the Riemann sums converge to the integrals: $c_n\to\int_0^1\frac{x}{1+x^{2}}dx=\frac12\ln2$ and $a_n\to\int_0^1\frac{dx}{1+x^{2}}=\frac\pi4$. The limit of the ratio is therefore $\frac{\frac12\ln2}{\frac\pi4}=\frac{2\ln2}{\pi}\approx0.4413$, which agrees with direct summation for $n=10^{5}$. Option (A) is half the correct value, option (C) is the reciprocal, and option (D) is twice the correct value.