Sum the following series to infinity : `tan^(-1)1/(1+1+1^2)+tan^(-1)1/(1+2+2^2)+tan^(-1)1/(1+3+3^2)+\ ddot`

Options

A. `(pi)/(4)`
B. `(pi)/(2)`
C. `(pi)/(3)`
D. `(pi)/(6)`

Answer

A

Explanation

Given series can be rewritten as
`underset(r=1)overset(oo)(sum)tan^(-1)((1)/(1+r+r^(2)))`
Now, `tan^(-1)((1)/(1+r+r^(2)))`
`=tan^(-1)((r+1-r)/(1+r(r+1)))`
`=tan^(-1)(r+1)-tan^(-1)(r)`
`therefore underset(r=1)overset(n)(sum)[tan^(-1)(r+1)-tan^-1r]`
`=tan^(-1)(n+1)-tan^(-1)(1)`
`=tan^(-1)(n+1)-(pi)/(4)`
`impliesunderset(r=1)overset(oo)(sum)tan^(-1)((1)/(1+r+r^(2)))`
`=(pi)/(2)-(pi)/(4)=(pi)/(4)`