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

1 Answer
Cesareo R.
Nov 5, 2016

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

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