The limit containing sum of arccosines DOES exist I'm trying to handle a quite complicated limit involving series. Such limits are really scary for me because I do not know any technique to compute them. The initial limit is $$\lim_{m\to\infty}\left(\ln\left(m+1+\sqrt{m^2+2m}\right)+\sum_{n=1}^{m+1}\arccos\frac1n-(m+1)\arccos\frac1{m+1}\right)$$
I'm thinking that determining of asymptotic behaviour of the sum $\sum_{n=1}^m\arccos\frac1n$ as $m\to\infty$ will help to evaluate this limit. Thank you for any contribution.
As an addition. Maybe it could be helpful. The last two terms I got after simplifying sum $$-\sum_{n=1}^m n\left(\arccos\frac1{n+1}-\arccos\frac1n\right)$$
Update. The limit could be rewritten (after changing $m+1\to m$) as $$\lim_{m\to\infty}\left(\ln\left(m+\sqrt{m^2-1}\right)+\sum_{n=1}^{m}\arccos\frac1n-m\arccos\frac1{m}\right)$$ or $$\lim_{m\to\infty}\left(\operatorname{arccosh}m+\sum_{n=1}^{m}\arccos\frac1n-m\arccos\frac1{m}\right)$$
The limit seems to exist. After computing numerically it seems to tend to $\color{red}{0.508132}$.
Update #2. After using Euler–Maclaurin formula I've gotten that $\sum_{n=1}^m\arccos\frac1n$ could be expressed as follows (for certain $p$, $f(x)=\arccos\frac1x$):
$$\begin{aligned}\sum_{n=1}^m\arccos\frac1n&=\int_1^m\arccos\frac1x\,dx+\frac12\arccos\frac1m+\sum_{k=1}^{\lfloor p/2\rfloor}\frac{B_{2k}}{(2k)!}\left(f^{(2k-1)}(m)-f^{(2k-1)}(1)\right)+R_p\\
&=m\arccos\frac1m-\operatorname{arccosh}m+\frac12\arccos\frac1m+\sum_{k=1}^{\lfloor p/2\rfloor}\frac{B_{2k}}{(2k)!}\left(f^{(2k-1)}(m)-f^{(2k-1)}(1)\right)+R_p
\end{aligned}$$
Thus $$\sum_{n=1}^m\arccos\frac1n-m\arccos\frac1m+\operatorname{arccosh}m=\frac12\arccos\frac1m+\sum_{k=1}^{\lfloor p/2\rfloor}\frac{B_{2k}}{(2k)!}\left(f^{(2k-1)}(m)-f^{(2k-1)}(1)\right)+R_p$$ and $$\lim_{m\to\infty}\left(\operatorname{arccosh}m+\sum_{n=1}^{m}\arccos\frac1n-m\arccos\frac1{m}\right)=\lim_{m\to\infty}\left(\frac12\arccos\frac1m+\sum_{k=1}^{\lfloor p/2\rfloor}\frac{B_{2k}}{(2k)!}\left(f^{(2k-1)}(m)-f^{(2k-1)}(1)\right)+R_p\right)$$
And I'm stuck. I do not know what I can do with these Bernoulli numbers and so on.
 A: The Euler-Maclaurin formula won't help here, instead you should use Abel's summation formula, that is
$$\sum_{1 \le n \le x} a_n f(n) = f(x) A(x) - \int_1^x A(t) f'(t) \ dt,$$
where $A(x) = \sum_{1 \le n \le x} a_n$. Apply this formula with $a_n =1$ and $f(x) = \arccos(1/x)$ and get
$$\tag{1}\sum_{k=1}^{m} \arccos(1/k) = m \arccos(1/m) - \int_1^m \lfloor x \rfloor \frac{1}{x \sqrt{x^2-1}} d x.$$
Note that $\arccos(1/m) \rightarrow \pi/2$ and $x (\arccos(1/x)-\arccos(1/(x+1)) \rightarrow 0$ if $x \rightarrow \infty$.  Thus the first term on the right-side in (1) cancels with the last term in your formula.
Additionally note that $g(x) := \lfloor x \rfloor- x$ is a bounded function. Thus
$$\int_1^\infty g(x) \frac{1}{x \sqrt{x^2-1}} \, dx $$
exists as a Lebesgue-integral (i.e. is absolutely integrable). This shows that we can replace $\lfloor x \rfloor$ by $x$. Therefore we have to calculate
$$ \int_1^m \frac{1}{\sqrt{x^2-1}} dx  =  \log(\sqrt{x^2-1}+x) \Big|_{x=1}^m = \log(\sqrt{m^2-1}+m).$$
We see that 
\begin{align}
&\lim_{m \rightarrow \infty} \left(\sum_{k=1}^m \arccos(1/k) - m \arccos(1/m) +\log(\sqrt{m^2-1}+m) \right) \\ &= \int_1^\infty (x - \lfloor x \rfloor) \frac{1}{x \sqrt{x^2-1}} dx.
\end{align}
Note that $\log(m+1 +\sqrt{m^2+2m}) -\log(\sqrt{m^2-1}+m) \rightarrow 0$ for $m \rightarrow \infty$.
