Lim n→∞Σk=1^2n(k)/(n^2+k^2 equals

Divide numerator and denominator by $n^2$: $$\frac{k}{n^2+k^2}=\frac1n\cdot\frac{k/n}{1+(k/n)^2}.$$ So the sum is a Riemann sum for $f(x)=\dfrac{x}{1+x^2}$ with $x=k/n$ running up to $2n/n=2$. Hence $$\lim=\int_0^{2}\frac{x}{1+x^2}\,dx=\frac12\ln(1+x^2)\Big|_0^2=\frac12\ln5.$$ The upper limit is $2$, not $1$, because $k$ goes to $2n$; using $1$ gives the distractor $\tfrac12\ln2$. The answer is a natural logarithm because the integrand is of the form $f'/f$ up to a constant.