解析してます
Sn={1+1/3+1/9+…+(1/3)^(n-1)}-{i+i/2+i/4+…+i(1/2)^(n-1)}
={ (1-(1/3)^(n-1))/(1-1/3) } + i{ (1-(1/3)^(n-1))/(1-1/3) }
→3/2 + 2i (n→∞)
はぃ、つぎぃ!!!
Sn={1+1/3+1/9+…+(1/3)^(n-1)}-{i+i/2+i/4+…+i(1/2)^(n-1)}
={ (1-(1/3)^(n-1))/(1-1/3) } + i{ (1-(1/3)^(n-1))/(1-1/3) }
→3/2 + 2i (n→∞)
はぃ、つぎぃ!!!