Prove that `tan^(- 1)(1/3)+tan^(- 1)(1/7)+tan^(- 1)(1/13)+..........+tan^-1 (1/(n^2+n+1))+......oo =pi/4`

Explanation

Let `P(n):tan^(-1)((1)/(3))+tan^(-1)((1)/(7))+…..+tan^(-1)((1)/(n^+n+1))=tan^(-1)((n)/(n+2))` ……..(i)
Step I For `n=1`
LHS of Eq. (i) `=tan^(-1)((1)/(3)) =tan^(-1)((1)/(1+2))=RHS of Eq. (i)
Therefore , P(1) is true .
Step II Assume that P(k) is true. Then ,
`P(k):tan^(-1)((1)/(3))+tan^(-1)((1)/(3))+tan^(-1)((1)/(7))+........+tan^(-1)((1)/(k^2+k+1))=tan^(-1)((k)/(k+2))`
Step III For `n=k+1`
`P(k+1):tan^(-1)((1)/(3))+tan^(-1)((1)/(7))+....+tan^(-1)((1)/(k^2+k+1))+tan^(-1)((1)/((k+1)^2+(k+1)+1))=tan^(-1)((k+1)/(k+3))`......(ii)
LHS of Eq. (ii)
`=tan^(-1)((1)/(3))+tan^(-1)((1)/(7))+.....+tan^(-1)((1)/(k^2+k+1))+tan^(-1)((1)/((k+1)^2+(k+1)+1))`
`=tan^(-1)((k+1)/(k+2))+tan^(-1)((1)/((k+1)^2+(k+1)+1))` [by aaumption atep]
`=tan^(-1)((k)/(1+(k+1)))+tan^(-1)((1)/(k^2+3k+3))`
`=tan^(-1)((k)/(1+(k+1)))+tan^(-1)((1)/(1+(k+1)(k+2)))`
`=tan^(-1)(((k+1)-1)/(1+(k+1).1))+tan^(-1)(((k+2)-(k+1))/(1+(k+2)(k+1)))`
`=tan^(-1)(k+1)-tan^(-1)1+tan^(-1)(k+2)-tan^(-1)(k+1)=tan^(-1)(k+2)-tan^(-1)1`
`=tan^(-1)((k+2-1)/(1+(k+2).1))=tan^(-1)((k+1)/(k+3))=` RHS of Eq. (ii)
This shows that the result is true for `n=k+1`. Hence. by the principle of mathematical induction . the result is true for all `n in N`.