Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics II / 7 / b

SetThe points strictly increase from zero to $x_{n,n}=n/(n+1)$. With $f(t)=\sin t$, the given expression is exactly the midpoint Riemann sumfor the partition of an interval ${\mathcal{P}}_n=(x_{0,n},\dots ,x_{n,n})$ of $[0,n/(n+1)]$. The sine argument is precisely the midpoint tag, since it is half the sum of the consecutive endpoints.

For a fixed integration interval $[0,1]$, append $x_{n+1,n}=1$ and choose the final tag ${\xi }_{n,n}=1$. The resulting Riemann sum $T_n$ satisfiesThis makes explicit the small final interval that the original sum omits.

We verify that the augmented partition of an interval in part i has vanishing mesh of a partition. Define ${\varphi }_n(t)=t^2/(t^2+n)$ for $t\ge 0$. Its derivative isSince $2u\le 1+u^2$ for $u\ge 0$, we have $0\le {\varphi }_n^{\prime }(t)\le 1/\sqrt{n}$. The mean value theorem therefore givesThe appended interval has length $1/(n+1)$, so $\Vert {\mathcal{P}}_n\Vert \le \max \{n^{-1/2},(n+1{)}^{-1}\}\to 0$ for the augmented partition of an interval.

The sine is continuous, hence Riemann integrable on $[0,1]$. By the Riemann sum convergence theorem, $T_n\to {\int }_0^1\sin tdt=1-\cos 1$, using the fundamental theorem of calculus and the cosine antiderivative. As $T_n-S_n=\sin 1/(n+1)\to 0$, the answer is