`sum_0^oo(1/3)^x=?`

sum_(0)^(oo)(1)/((3n+1)(3n+2))

sum_(j=0)^(oo)(-3)^(-j)

Let x= sum_(n=0)^oo (-1)^n (tantheta)^(2n) and y = sum_(n=0)^oo (costheta)^(2n) qhere theta in (0,pi/4) , then

Let x= sum_(n=0)^oo (-1)^n (tantheta)^(2n) and y = sum_(n=0)^oo (costheta)^(2n) qhere theta in (0,pi/4) , then

If a=sum_(n=0)^(oo)x^(n),b=sum_(n=0)^(oo)y^(n),c=sum_(n=0)^(oo)(xy)^(n) where |x|,|y|<1 then

If a=sum_(n=0)^(oo)x^(n),b=sum_(n=0)^(oo)y^(n),c=sum_(n=0)^(oo)(xy)^(n) where |x|,|y|<1 then

p=sum_(n=0)^(oo) (x^(3n))/((3n)!) , q=sum_(n=1)^(oo) (x^(3n-2))/((3n-2)!), r = sum_(n=1)^(oo) (x^(3n-1))/((3n-1)!) then p + q + r =

Find the value of sum_(i=0)^(oo)sum_(j=0)^(oo)sum_(k=0)^(oo)(1)/(3^(i)3^(j)3^(k)) .