Find the `sum_(k=1)^(oo) sum_(n=1)^(oo)k/(2^(n+k))`.

Explanation

`S=sum_(k=1)^(oo)sum_(n=1)^(oo)k/(2^(n+k))`
`=sum_(k=1)^(oo)sum_(n=1)^(oo)k/(2^(k))cdot1/(2^(n))`
`=sum_(k=1)^(oo)(k/2^(k)sum_(n=1)^(oo)1/2^(n))`
`=sum_(k=1)^(oo)k/(2^(k))(1/2+1/2^(2)+1/2^(3)+…oo)`
`=sum_(k=1)^(oo)k/2^(k)(1/2)/(1-1/2)`
`=sum_(k=1)^(oo)k/2^(k)`
`thereforeS=1/2+2/2^(2)+3/2^(3)....` (1)
`thereforeS/2=1/2^(2)+2/2^(3)+3/2^(4)+...` (2)
Subtracting (2) from (1), we get
`S/2=1/2+1/2+1/2^(3)+...=1`
`thereforeS=2`