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

 A: A more convenient way to state the sequence is:
$$
\frac{\sum_{k=1}^n\cos\frac{1}{k}\arccos\frac{1}{n-k+1}}{n}
$$
Note that for $0\leq x\leq 1$ we have $1-x\leq\cos x\leq 1$ and $\frac{\pi}{2}-\frac{\pi}{2}x\leq\arccos x\leq \frac{\pi}{2}-x$. Therefore, we have
$$
\tfrac{\pi}{2}(1-\tfrac{1}{k})(1-\tfrac{1}{n-k+1})\leq\cos\frac{1}{k}\arccos\frac{1}{n-k+1}\leq\frac{\pi}{2}-\frac{1}{n-k+1}.
$$
Thus, a lower bound for the limit is given by the sequence
$$
\frac{\sum_{k=1}^n\frac{\pi}{2}(1-\tfrac{1}{k})(1-\tfrac{1}{n-k+1})}{n}=\frac{\pi}{2}-\frac{\sum_{k=0}^n\frac{(n-k+1)+k-1}{k(n-k+1)}}{n}=\frac{\pi}{2}-\sum_{k=1}^n\frac{1}{k(n-k+1)}
$$
The terms in the latter sum can be rewritten to
$$
\frac{1}{k(n+1)}+\frac{1}{(n-k+1)(n+1)}
$$
so we get the sums
$$
\frac{1}{n+1}\sum_{k=1}^n\frac{1}{k}\qquad\text{and}\qquad\frac{1}{n+1}\sum_{k=1}^n\frac{1}{n-k+1}.
$$
Sacha indicated a proof in the comments that these sums tend to zero. Similarly, for the upper bound we have
$$
\frac{\sum_{k=1}^n\frac{\pi}{2}-\tfrac{1}{n-k+1}}{n}=\frac{\pi}{2}-\frac{\sum_{k=1}^n\frac{1}{n-k+1}}{n}\to\frac{\pi}{2}
$$
This gives an upper bound of $\frac{\pi}{2}$. This finishes the proof that the limit is $\frac{\pi}{2}$.

Note: Another proof that $\frac{1}{n}\sum_{k=1}^n\frac{1}{k}\to 0$. Using Cauchy-Schwarz, we see that
$$
\sum_{k=1}^n\frac{1}{k}\leq \sqrt{n}\sqrt{\textstyle\sum_{k=1}^n\tfrac{1}{k^2}}
$$
Therefore we get
$$
\frac{1}{n}\sum_{k=1}^n\frac{1}{k}\leq\sqrt{\frac{\textstyle\sum_{k=1}^n\tfrac{1}{k^2}}{n}}
$$
In the square root on the right hand side, the numerator converges, so the whole tends to zero.
A: What follows is a little hand-wavy, and I wish I had more rigorous demonstration, but the post is too big for a comment.
$$ \begin{eqnarray}
   \frac{1}{n} \sum_{k=1}^n \cos\left(\frac{1}{k}\right) \arccos\left( \frac{1}{n+1-k} \right) &=& \frac{1}{n} \sum_{k=1}^n \left( 1 - 2 \sin^2\left(\frac{1}{2 k}\right) \right) \left( \frac{\pi}{2} - \arcsin\left( \frac{1}{n+1-k} \right) \right) 
\end{eqnarray}
$$
We can now split the sum in two parts, $1 \leqslant k \leqslant \lfloor\frac{n}{2}\rfloor$ and $\lfloor\frac{n}{2}\rfloor < k \leqslant n$. In each of these parts, either $\sin$, or $\arcsin$ will be small, and in the limiting value will be $\frac{\pi}{2}$:
In[28]:= Table[
  N[1/n Sum[Cos[1/k] ArcCos[1/(n - k + 1)], {k, 1, n}], 
   50], {n, {5000, 10000, 100000}}] // N

Out[28]= {1.56861, 1.56963, 1.57066}


More rigorously:
$$
\frac{1}{n} \sum_{k=1}^n \left( 1 - 2 \sin^2\left(\frac{1}{2 k}\right) \right) \left( \frac{\pi}{2} - \arcsin\left( \frac{1}{n+1-k} \right) \right) = \frac{\pi}{2} - \frac{\pi}{n} \sum_{k=1}^n \sin^2 \frac{1}{k} - \frac{1}{n} \sum_{k=1}^n \arcsin\frac{1}{k} + \frac{2}{n} \sum_{k=1}^n \sin^2\left( \frac{1}{k}\right) \arcsin\left(\frac{1}{n+1-k}\right)
$$
Now:
$$
   0 \leqslant \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \sin^2\left(\frac{1}{k}\right) \leqslant
   \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{k^2} = 0
$$
$$
   0 \leqslant \lim_{n\to \infty} \frac{1}{n} \sum_{k=1}^n \arcsin\left(\frac{1}{k}\right) \leqslant \lim_{n \to \infty} \frac{\pi}{2n} \sum_{k=1}^n \frac{1}{k} = \lim_{n \to \infty} \frac{\pi}{2 n} \ln(n) = 0
$$
$$
  0 \leqslant \lim_{n \to \infty} \frac{2}{n} \sum_{k=1}^n \sin^2 \left( \frac{1}{k} \right) \arcsin\left(\frac{1}{n+1-k} \right) \leqslant \lim_{n \to \infty} \frac{2}{n} \sum_{k=1}^n \frac{1}{k^2} \frac{1}{n+1-k} \leqslant \lim_{n \to \infty} \frac{2}{n} \sum_{k=1}^n \frac{1}{k^2} = 0
$$
A: 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.
