The sum of the infinite terms of the series ` "tan"^(-1)((1)/(3))+ "tan"^(-1)((2)/(9)) + tan^(-1)((4)/(33)) + .... ` is equal to ` (pi)/(n)` The value of n is:

"tan"^(-1)((1)/(3))+ "tan"^(-1)((2)/(9)) + tan^(-1)((4)/(3^(3))) + ....oo is equal to

tan^(-1)((3)/(n))+tan^(-1)((4)/(n))=(pi)/(2)

Find the sum to infinite terms of the series tan^(-1)((1)/(3))+tan^(-1)((2)/(9))+"........."+tan^(-1)((2^(n-1))/(1+2^(2n-1)))+"......" .

Prove that, tan^(-1) (4/3) + tan^(-1) (12/5) = pi - tan^(-1)(56/33) .

If "S" is the sum of the first "10" terms of the series tan "^(-1)((1)/(3))+tan^(-1)((1)/(7))+tan^(-1)((1)/(13))+tan^(-1)((1)/(21))+.... ,then "tan(S)" is equal to:

The sum to infinite terms of the series tan^(-1)((2)/(1-1^(2)+1^(4)))+tan^(-1)((4)/(1-2^(2)+2^(4)))+tan^(-1)((6)/(1-3^(2)+3^(4)))+

The value of tan^(-1)((1)/(3))+tan^(-1)((2)/(9))+tan^(-1)((4)/(33))+tan^(-1)((8)/(129))+...n terms is:

Find the sum of the series : (tan^(-1))(1)/(3)+(tan^(-1))(2)/(9)+...+(tan^(-1))(2^(n-1))/(1+2^(2n-1))+......oo

tan^(-1)((n)/(n+1))-tan^(-1)(2n+1)=(3 pi)/(4)