2017
Problem - 4558
For $n=1, 2,\dots$, let $x_n=\displaystyle\sum_{k=n+1}^{9n}\frac{k}{9n^2 + k^2}$. Find the value of $\displaystyle\lim_{n\to\infty}x_n$.
Note that $x_n$ can be rewritten as
$$x_n =\sum_{k=n+1}^{9n}\frac{k}{9n^2+k^2}=\sum_{k=n+1}^{9n}\frac{1}{n}\cdot\frac{k/n}{9+(k/n)^2}$$
which is the $n^{th}$ Riemann sum for, and thus, converge to
$$\int_1^9\frac{x}{9+x^2}dx=\frac{1}{2}\ln(9+x^2)|_1^9=\boxed{\ln 3}$$