Find the sum `cot^(-1) 2 + cot^(-1) 8 + cot^(-1) 18 + ...oo`

Answer

`(pi)/(4)`

Explanation

Let `r_(n)` denotes the `nth` term of the series. Then
`t_(n) = cot^(-1) 2n^(2)`
`= tan^(-1).(1)/(2n^(2))`
`= tan^(-1).((2n + 1) - (2n -1))/(1 + (4n^(2) -1))`
`= tan^(-1) (2n +1) - tan^(-1) (2n -1)`...(i)
Putting `n = 1, 2, 3`,.., etc. in (i), we get
`t_(1) = tan^(-1) 3 - tan^(-1) 1`
`t_(2) = tan^(-1) 5 - tan^(-1) 3`
`{:(t_(3) = tan^(-1) 7 - tan^(-1) 5 ),(vdots),(t_(n) = tan^(-1) (2n + 1) - tan^(-1) (2n -1)):}`
Adding, we get
`S_(n) = tan^(-1) (2n + 1) - tan^(-1) 1`
as `n rarr oo, tan^(-1) (2n + 1) rarr pi//2`
Hence, the required sum is `(pi)/(4)`