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.

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    $\begingroup$ It was a long time I calculated things like this. But maybe you can estimate this sum with integrals of some kind. $\endgroup$ – mathreadler Nov 22 '18 at 13:16
  • $\begingroup$ Have you tried Euler-Maclaurin formula? $\endgroup$ – Szeto Nov 22 '18 at 14:22
  • $\begingroup$ No, I haven't. I supposed that it would not help because I would have to find n-th derivative of $\arccos\frac1x$ which (I think) hasn't closed form $\endgroup$ – Mikalai Parshutsich Nov 22 '18 at 15:49
  • $\begingroup$ Please have titles that reflect the content of your question, rather announcements that would only be understood after spending some nontrivial amount of time reading through your question. $\endgroup$ – Asaf Karagila Nov 27 '18 at 6:45

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 $m \rightarrow \infty$.

  • $\begingroup$ Thank you very much for showing that wonderful technique of calculating infinite sums. But I have some doubt: don't we have to add $\log$ under the limit in the last equation? $\endgroup$ – Mikalai Parshutsich Nov 23 '18 at 5:11
  • $\begingroup$ ... you've edited typos almost everywhere. But there is one left in the last equation: $\sqrt{x^2-1}$ should be in the denominator under integral instead of $\sqrt{1-x}$ $\endgroup$ – Mikalai Parshutsich Nov 23 '18 at 5:26
  • $\begingroup$ And after all... the last integral is the very on which I tried to compute in the first place :'-( $\endgroup$ – Mikalai Parshutsich Nov 23 '18 at 5:43
  • $\begingroup$ Yes, there was a sign-error! Hopefully, I have corrected all errors. :-) A similar argument can be also used in order to prove that $\lim_{n \rightarrow \infty} ( \sum_{k=1}^n \frac{1}{k}-\ln(n))$ exists and has an integral representation. (This limes is known as the 'Euler–Mascheroni constant'.) $\endgroup$ – p4sch Nov 23 '18 at 8:59

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