Sigma k=1~無限大 k*(2/3)^k 請問怎麼算?

設 S=1*(2/3)+2*(2/3)^2+.......+ k*(2/3)^k+...... .....(1)
(2/3)S=1*(2/3)^2+2*(2/3)^3+.... k*(2/3)^(k+1)+.... ....(2)
(1)-(2)
(1/3)S=(2/3)+(2/3)^2+........+(2/3)^k+(2/3)^(k+1)+....
=(2/3)/[1-(2/3)]
=2
S=6 .....Ans

Sol
A=Σ(k=1 to ∞)_k*(2/3)^k
lim(n->∞)_k*(2/3)^n=0
A=1*(2/3)^1+2*(2/3)^2+3*(2/3)^3+4*(2/3)^4+…
2A/3= 1*(2/3)^2+2*(2/3)^2+3*(2/3)^3+…
A/3=(2/3)^1+(2/3)^2+(2/3)^3+(2/3)^4+…=1/(1-2/3)=3
A=9