`sum_(i=0)^oosum_(j=0)^oosum_(k=0)^oo1/(3^i3^i3^k)`

Explanation

Let `S=sum_(i=0)^(oo)sum_(j=0)^(oo)sum_(k=0)^(oo)(1)/(3^(i)3^(j)3^(k)) " " [i nejnek]`
We will first of all find the sum without any rsetriction on `I,j,k`
Let `S_(1)=sum_(i=0)^(oo)sum_(j=0)^(oo)sum_(k=0)^(oo)(1)/(3^(i)3^(j)3^(k))=(sum_(i=0)^(oo)(1)/(3^(i)))^(3)`
`((3)/(2))^(3)=(27)/(8)`
Case I If `i=j=k`
Let `S_(2)=sum_(i=0)^(oo)sum_(j=0)^(oo)sum_(k=0)^(oo)(1)/(3^(i)3^(j)3^(k))`
`=sum_(i=0)^(oo)(1)/(3^(3i))=1+(1)/(3^(3))+(1)/(3^(6))+"..."=(1)/(1-(1)/(3^(3)))=(27)/(26)`
Case II If `i=j=k`
Let `S_(3)=sum_(i=0)^(oo)sum_(j=0)^(oo)sum_(k=0)^(oo)(1)/(3^(i)3^(j)3^(k))=(sum_(i=0)^(oo)(1)/(3^(2i)))(sum_(k=0)^(oo)(1)/(3^(k)))" " [:.knei}`
`=sum_(i=0)^(oo)(1)/(3^(2i))((3)/(2)=(1)/(3^(i)))sum_(i=0)^(oo)(3)/(2)*(1)/(3^(2i))-sum_(i=0)^(oo)(1)/(3^(3i))`
`=(3)/(2)*(9)/(8)-(27)/(26)=(135)/(208)`
Hence required sum, `S=S_(1)-S_(2)-3S_(3)`
`=(27)/(8)-(27)/(26)-3((135)/(208))=(27xx26-27xx8-3xx135)/(208)=(81)/(208)`.