Difference between revisions of "Coalescence"
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<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>\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> | + | <math>4N\sum_{i=2}^\infty\frac{1}{n(n-1)}=4N\sum_{i=1}^\infty\frac{1}{n(n+1)}=4N\sum_{i=1}^\infty\frac{1}{n}-\frac{1}{n+1}</math> |
Revision as of 01:46, 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]
[math]4N\sum_{i=2}^\infty\frac{1}{n(n-1)}=4N\sum_{i=1}^\infty\frac{1}{n(n+1)}=4N\sum_{i=1}^\infty\frac{1}{n}-\frac{1}{n+1}[/math]
