Clearly, \begin{align}&\frac1{1\cdot2}+\frac1{2\cdot3}+\ldots+\frac1{(n-1)n}+\frac1{n\cdot1} \\=& 1-\frac12+\frac12-\frac13+\ldots+\frac1{n-1}-\frac1n+\frac1n=1. \end{align} See also my related conjectures available from http://oeis.org/A322069 and http://oeis.org/A322070.

Clearly, \begin{align}&\frac1{1\cdot2}+\frac1{2\cdot3}+\ldots+\frac1{(n-1)n}+\frac1{n\cdot1} \\=& 1-\frac12+\frac12-\frac13+\ldots+\frac1{n-1}-\frac1n+\frac1n=1. \end{align} If $n=2k+1$, then we also have \begin{align}&\frac1{1\cdot2}+\frac1{2\cdot3}+\ldots+\frac1{k(2k+1)}+\frac1{(2k+1)(k+1)} \\&+\frac1{(k+1)(k+2)}+\cdots+\frac1{(2k-1)2k}+\frac1{2k\cdot1} \\=&1.\end{align} See also my related conjectures available from http://oeis.org/A322069 and http://oeis.org/A322070.