Coalescence

Sum of the Infinite Series

[math]\sum_{i=2}^\infty\frac{2N}{\frac{n(n-1)}{2}}=\sum_{i=2}^\infty\frac{4N}{n(n-1)}=4N\sum_{i=2}^\infty\frac{1}{n(n-1)}[/math]

[math]4N\sum_{i=2}^\infty\frac{1}{n(n-1)}=4N\sum_{i=1}^\infty\frac{1}{n(n+1)}[/math]