Question

`sum_(k=1)^oo tan^(-1)(1/(2k^2))=` ?

Answer 1

`pi/4`

Explanation:

`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`