Find probability generating function when P(Y=r)=kP(X=r)

Question

The variable Y can take only the values 1,2,3,... and is such that P(Y=r)=kP(X=r), where X~Po($\lambda$). Show that the probability generating function of Y is given by: $$G_Y(t)=\frac{e^{\lambda t}-1}{e^\lambda-1}$$

\begin{align} G_Y(t)&=\sum_{r=1}^\infty t^rP(Y=r)\\ &=\sum_{r=1}^\infty t^rkP(X=r)\\ &=\sum_{r=1}^\infty t^rk\frac{e^{- \lambda}\lambda^r}{r!}\\ &=ke^{-\lambda}(\frac{t\lambda}{1!}+\frac{(t\lambda)^2}{2!}+...)\\ &=ke^{-\lambda}(e^{\lambda t}-1) \end{align} How is the k eliminated in the final equation?

Answer 1

$ 1=\sum\limits_{r=1}^{\infty} P(Y=r)=k\sum\limits_{r=1}^{\infty} P(X=r)=k(1-P(X=0))=k(1-e^{-\lambda})$. So $k=\frac 1 {1-e^{-\lambda}}$.