#sum_(k=1)^oo cot^(-1)(2k^2)=sum_(k=1)^oo tan^(-1)(1/(2k^2))=pi/4#
We know that
#tan^(-1)a-tan^(-1)b=tan^(-1)((a-b)/(1+ab))# so taking
#a=2k+1# and #b=2k-1# we have
#tan^(-1)(1/(2k^2))=tan^(-1)(2k+1)-tan^(-1)(2k-1)# so
#sum_(k=1)^oo tan^(-1)(1/(2k^2))=sum_(k=1)^oo(tan^(-1)(2k+1)-tan^(-1)(2k-1))# which is known as a telescopic series
so we have
#{(k=1->tan^(-1)(1/2)=tan^(-1)(3)-tan^(-1)(1)),(k=2->tan^(-1)(1/8)=tan^(-1)(5)-tan^(-1)(3)),(k=3->tan^(-1)(1/18)=tan^(-1)(7)-tan^(-1)(5)),(k=4->tan^(-1)(1/32)=tan^(-1)(9)-tan^(-1)(7)),(cdots):}#
Summing up each term we have
#sum_(k=1)^oo tan^(-1)(1/(2k^2))=tan^(-1)(0)-tan^(-1)(1)=pi/2-pi/4=pi/4#