Revision as of 02:01, 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]?

Multiply both sides by one to equalize the denominators and combine.

[math]\frac{1}{i}-\frac{1}{i+1}=\frac{i+1}{i+1}\frac{1}{i}-\frac{i}{i}\frac{1}{i+1}=\frac{i+1-i}{i(i+1)}=\frac{1}{i(i+1)}[/math]

Plug in the first few numbers of the sum to see the pattern.

[math]\sum_{i=1}^\infty\frac{1}{i}-\frac{1}{i+1} = \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + \cdots[/math]

After the first one the pairs of fractions cancel out: +1/2 -1/2, +1/3, -1/3, +1/4, -1/4, ... this pattern continues to infinity. So,

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

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