= Solution
{c}
Set
$$
x_{k,n}=\frac{k^2}{k^2+n}\quad(0\leq k\leq n),\qquad
\xi_{k,n}=\frac{x_{k,n}+x_{k+1,n}}2\quad(0\leq k<n).
$$
The 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 sum>
$$
\boxed{S_n=\sum_{k=0}^{n-1}(x_{k+1,n}-x_{k,n})f(\xi_{k,n})}
$$
for the <partition of an interval> $\mathcal P_n=(x_{0,n},\ldots,x_{n,n})$ 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> $T_n$ satisfies
$$
\boxed{T_n=S_n+\frac{\sin1}{n+1}.}
$$
This makes explicit the small final <interval> that the original sum omits.
Back to article page
