Difference between revisions of "Coalescence"
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| − | <math>\frac{2N}{\frac{n(n-1)}{2}}=\frac{4N}{n(n-1)}</math> | + | <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> |
Revision as of 01:30, 18 September 2018
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]
