TITLE: $\cos(x)$ and $\arccos(x)$ couple limit
QUESTION [11 upvotes]: Find the value of the following limit:
$$\lim_{n\to\infty} \frac {\cos 1 \cdot \arccos \frac{1}{n}+\cos\frac
{1}{2} \cdot \arccos \frac{1}{(n-1)}+ \cdots +\cos \frac{1}{n} \cdot
 \arccos{1}}{n}$$

REPLY [4 votes]: This is too long for a comment, so I am posting it as an answer, although it doesn't completely resolve the question.
I shall prove that $\frac{\pi}{2}$ is an upper bound.
Lemma. For all $n\in\Bbb N$, we have $\frac{1}{n} \sum\limits_{k=1}^n \cos\left(\frac{1}{k}\right) \arccos\left( \frac{1}{n+1-k} \right) \leq \frac{\pi}{2}$.
Proof. Since both functions are decreasing, we have: $$\frac{1}{n} \sum_{k=1}^n \cos\left(\frac{1}{k}\right) \arccos\left( \frac{1}{n+1-k} \right) \leq \frac1nn\cos(0)\arccos(0)=\frac{\pi}{2},$$
which shows the desired inequality. $\square$
As Sasha mentions above this is probably also the value of the limit.
As Chris notes in the comment below this answer, it is possible to prove that $\frac{\pi}2$ is also a lower bound by a simple application of AM-GM inequality and the Cesaro-Stolz theorem, completing the proof.