Difference between revisions of "Coalescence"
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=Sum of the Infinite Series= | =Sum of the Infinite Series= | ||
<math>\sum_{i=2}^\infty\frac{2N}{\frac{i(i-1)}{2}}=\sum_{i=2}^\infty\frac{4N}{i(i-1)}=4N\sum_{i=2}^\infty\frac{1}{i(i-1)}</math> | <math>\sum_{i=2}^\infty\frac{2N}{\frac{i(i-1)}{2}}=\sum_{i=2}^\infty\frac{4N}{i(i-1)}=4N\sum_{i=2}^\infty\frac{1}{i(i-1)}</math> | ||
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| + | Note shifting the index starting point down by one, i=1 instead of i=2 in the sum. | ||
<math>4N\sum_{i=2}^\infty\frac{1}{i(i-1)}=4N\sum_{i=1}^\infty\frac{1}{i(i+1)}=4N\sum_{i=1}^\infty\frac{1}{i}-\frac{1}{i+1}</math> | <math>4N\sum_{i=2}^\infty\frac{1}{i(i-1)}=4N\sum_{i=1}^\infty\frac{1}{i(i+1)}=4N\sum_{i=1}^\infty\frac{1}{i}-\frac{1}{i+1}</math> | ||
Revision as of 01:49, 18 September 2018
Sum of the Infinite Series
[math]\sum_{i=2}^\infty\frac{2N}{\frac{i(i-1)}{2}}=\sum_{i=2}^\infty\frac{4N}{i(i-1)}=4N\sum_{i=2}^\infty\frac{1}{i(i-1)}[/math]
Note shifting the index starting point down by one, i=1 instead of i=2 in the sum.
[math]4N\sum_{i=2}^\infty\frac{1}{i(i-1)}=4N\sum_{i=1}^\infty\frac{1}{i(i+1)}=4N\sum_{i=1}^\infty\frac{1}{i}-\frac{1}{i+1}[/math]
Why is
[math]\frac{1}{i(i+1)}=\frac{1}{i}-\frac{1}{i+1}[/math]?
