Question Number 104016 by Dwaipayan Shikari last updated on 18/Jul/20

Answered by OlafThorendsen last updated on 18/Jul/20

arctanx−arctany = arctan((x−y)/(1+xy))  arctan(2k+1)−arctan(2k−1) = arctan(((2k+1)−(2k−1))/(1+(2k+1)(2k−1)))  =  arctan(2/(4k^2 )) = arctan(1/(2k^2 ))  S_n  =Σ_(k=1) ^(k=n) arctan(1/(2k^2 ))  S_n  =Σ_(k=1) ^(k=n) arctan(2k+1)−arctan(2k−1)  S_n  = arctan3−arctan1  +arctan5−arctan3...  +arctan(2k+1)−arctan(2k−1)  S_n  = arctan(2k+1)−arctan1  lim_(n→∞) S_n  = (π/2)−(π/4) = (π/4)  Σ_(k=1) ^∞ arctan(1/(2k^2 )) = (π/4)