$\cos(x)$ 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 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.
