if `1+1/8+1/27+...........oo=x` then `1+1/27+1/125+............oo` equal to

1 ? 27 64 125

1 ? 27 64 125

If x=9^(1/3) 9^(1/9) 9^(1/27) ......oo y= 4^(1/3) 4^(-1/9) 4^(1/27) ...... oo and z= sum_(r=1)^oo (1+i)^-r then , the argument of the complex number w = x+yz is

If x=9^(1/3) 9^(1/9) 9^(1/27) ......ad inf y= 4^(1/3) 4^(-1/9) 4^(1/27) ...... ad inf and z= sum_(r=1)^oo (1+i)^-r then , the argument of the complex number w = x+yz is

If x=9^(1/3) 9^(1/9) 9^(1/27) ......ad inf y= 4^(1/3) 4^(-1/9) 4^(1/27) ...... ad inf and z= sum_(r=1)^oo (1+i)^-r then , the argument of the complex number w = x+yz is

If x=9^(1/3) 9^(1/9) 9^(1/27) ......and if y= 4^(1/3) 4^(-1/9) 4^(1/27) ...... and if z= sum_(r=1)^oo (1+i)^-r then , the argument of the complex number w = x+yz is

If y= x^(1/3).x^(1/9).x^(1/27)...oo then y=?

(1+1/(2!)+1/(4!)+1/(6!).....oo)/(1+1/(3!)+1/(5!)+1/(7!)+....oo) equals

lim _( x to oo) (1+ (1)/(x))^(x//2) is equal to-