\( \def \l {\left} \def \r {\right} \def \f {\frac} \def \b#1{\l(#1\r)} \def \root [#1]#2{\sqrt[\leftroot{2}\uproot{2}\scriptstyle #1]{#2}} \DeclareMathOperator{\acoth}{acoth} \)

Show that $\sum\limits_{i=2}^n \acoth i = \f{1}{2} \ln \f{n(n+1)}{2}$

Pub: Sep. 14, 2014 | Wri: unknown


\[ \text{Let } S_n = \sum\limits_{i=2}^n \acoth i \\[15pt] \acoth x = \f{1}{2} \ln \f{x+1}{x-1} \\[12pt] \begin{split} S_n &= \acoth 2 + \acoth 3 + \dots + \acoth n \\[6pt] &= \f{1}{2} \b{\ln \f{2+1}{2-1} + \ln \f{3+1}{3-1} + \dots + \ln \f{n+1}{n-1}} \\[6pt] &= \f{1}{2} \ln \f{(2+1)(3+1)\dots(n+1)}{(2-1)(3-1)\dots(n-1)} \\[6pt] &= \f{1}{2} \ln \f{\f{(n+1)!}{2}}{(n-1)!} \\[6pt] &= \f{1}{2} \ln \f{n(n+1)}{2} \end{split} \]