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