Interesting Limit (14)

Let f(x)=x^\sqrt{3} . The trapezoidal sums of $f(x)$ on $[0,1]$ are $T_n=-\frac{1}{2n}+\sum_{k=1}^n (\frac{k}{n})^{\sqrt3} \frac{1}{n}$ , since both endpoints "count half." The trapezoidal sum is a second-order method, so that $0=\lim_{n \to \infty} n\left(T_n-\int_{0}^{1} f(x)dx\right)=\lim_{n \to \infty}\left( -\frac{1}{2}+\sum_{k=1}^n \left(\frac{k}{n}\right)^{\sqrt3} -\frac{n}{1+\sqrt3}\right)=-\frac{1}{2}+\lim_{n \to \infty}\left( \sum_{k=1}^n \frac{k^{\sqrt3}}{n^{\sqrt3}} -\frac{n}{1+\sqrt3}\right)$ Thus the limit we seek is $\boxed{0.5}$ . (The answer comes out to be the same if we replace $\sqrt3$ by any other positive real number.)