\(
\def \l {\left}
\def \r {\right}
\def \f {\frac}
\def \b#1{\l(#1\r)}
\def \root [#1]#2{\sqrt[\leftroot{2}\uproot{2}\scriptstyle #1]{#2}}
\DeclareMathOperator{\acoth}{acoth}
\)
Show that $\sum\limits_{i=2}^n \acoth i = \f{1}{2} \ln \f{n(n+1)}{2}$
Pub: Sep. 14, 2014 | Wri: unknown
\[
\text{Let } S_n = \sum\limits_{i=2}^n \acoth i \\[15pt]
\acoth x = \f{1}{2} \ln \f{x+1}{x-1} \\[12pt]
\begin{split}
S_n &= \acoth 2 + \acoth 3 + \dots + \acoth n \\[6pt]
&= \f{1}{2} \b{\ln \f{2+1}{2-1} + \ln \f{3+1}{3-1} + \dots + \ln \f{n+1}{n-1}} \\[6pt]
&= \f{1}{2} \ln \f{(2+1)(3+1)\dots(n+1)}{(2-1)(3-1)\dots(n-1)} \\[6pt]
&= \f{1}{2} \ln \f{\f{(n+1)!}{2}}{(n-1)!} \\[6pt]
&= \f{1}{2} \ln \f{n(n+1)}{2}
\end{split}
\]