1991 TYS

Given that \({{u}_{n}}=\frac{{{2}^{n}}}{n}\), show that \({{u}_{n+1}}-{{u}_{n}}=\frac{{{2}^{n}}\left( n-1 \right)}{n\left( n+1 \right)}\) where \(n\) is a positive integer.

Hence, show that \(\sum\limits_{n=1}^{N}{\frac{{{2}^{n}}\left( n-1 \right)}{n\left( n+1 \right)}}=\frac{{{2}^{N+1}}}{N+1}-2\).