tan^{- 1}(1/3) +tan^{- 1}(1/7) + tan^{- 1}(1/13) +........................... to =/4

I need the steps for the above problem

The r-th term of the series is tan^-1[ 1 / (r²+r+1)] = tan^-1 [ (r+1-r)/1+r(r+1)] = tan^-1(r+1) - tan^-1(r).

So summation = [tan^-1(2) - tan^-1(1)] + [tan^-1(3) - tan^-1(2)] + ..... + [tan^-1(n+1) - tan-1(n)]

ie, Summation = tan^-1(n+1) - tan^-1(1)

As n tends to infinity ;

Summation = tan^-1(infinity + 1) - tan^-1(1) = pi/2 - pi/4 = pi/4