Revision as of 01:47, 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]

[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]

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 =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>\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>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>
+<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>