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 $\sum\limits_{n=1}^{N}{\frac{{{2}^{n}}\left( n-1 \right)}{n\left( n+1 \right)}}=\frac{{{2}^{N+1}}}{N+1}-2$.