Krdytkyu

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 $$lim_{mtoinfty}left(lnleft(m+1+sqrt{m^2+2m}right)+sum_{n=1}^{m+1}arccosfrac1n-(m+1)arccosfrac1{m+1}right)$$
I'm thinking that determining of asymptotic behaviour of the sum $sum_{n=1}^marccosfrac1n$ as $mtoinfty$ 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 $$-sum_{n=1}^m nleft(arccosfrac1{n+1}-arccosfrac1nright)$$

Update. The limit could be rewritten (after changing $m+1to m$) as $$lim_{mtoinfty}left(lnleft(m+sqrt{m^2-1}right)+sum_{n=1}^{m}arccosfrac1n-marccosfrac1{m}right)$$ or $$lim_{mtoinfty}left(operatorname{arccosh}m+sum_{n=1}^{m}arccosfrac1n-marccosfrac1{m}right)$$
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 $sum_{n=1}^marccosfrac1n$ could be expressed as follows (for certain $p$, $f(x)=arccosfrac1x$):

$$begin{aligned}sum_{n=1}^marccosfrac1n&=int_1^marccosfrac1x,dx+frac12arccosfrac1m+sum_{k=1}^{lfloor p/2rfloor}frac{B_{2k}}{(2k)!}left(f^{(2k-1)}(m)-f^{(2k-1)}(1)right)+R_p\
&=marccosfrac1m-operatorname{arccosh}m+frac12arccosfrac1m+sum_{k=1}^{lfloor p/2rfloor}frac{B_{2k}}{(2k)!}left(f^{(2k-1)}(m)-f^{(2k-1)}(1)right)+R_p
end{aligned}$$
Thus $$sum_{n=1}^marccosfrac1n-marccosfrac1m+operatorname{arccosh}m=frac12arccosfrac1m+sum_{k=1}^{lfloor p/2rfloor}frac{B_{2k}}{(2k)!}left(f^{(2k-1)}(m)-f^{(2k-1)}(1)right)+R_p$$ and $$lim_{mtoinfty}left(operatorname{arccosh}m+sum_{n=1}^{m}arccosfrac1n-marccosfrac1{m}right)=lim_{mtoinfty}left(frac12arccosfrac1m+sum_{k=1}^{lfloor p/2rfloor}frac{B_{2k}}{(2k)!}left(f^{(2k-1)}(m)-f^{(2k-1)}(1)right)+R_pright)$$
And I'm stuck. I do not know what I can do with these Bernoulli numbers and so on.

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 $m rightarrow infty$.