We can associate two $Q_p$ vector spaces to a p-divisible group, and I'm a little confused about the relation between these two groups. first of all, I think part of my problem is that when papers discuss these things, they claim that if $R$ is p-complete, sending $x\to p^nx$ is topologically nilpotent on $G(R)$, in particular, it is nilpotent if $p^nR=0$ I don't understand why this is true and I would appreciate an explanation or a reference.

But if we accept the claim it is easy to see that for any connected $R$ such that $p^nR=0$,we have: $$\tilde{G}(R):=\lim_{p:G\to G}(G(R))=Hom(\mathbb{Q}_p/\mathbb{Z}_p,G(R))[1/p]$$ on the other hand one has $$T_G(R):=\lim(G[n](R))=Hom(\mathbb{Q}_p/\mathbb{Z}_p,G(R))$$ So it seems to me that if $p^nR=0$, then $\tilde{G}(R)$ is just rational Tate module, and they are different only in mixed characteristic. Is my understanding correct?

We can associate two $\mathbb Q_p$ vector spaces to a $p$-divisible group, and I'm a little confused about the relation between these two groups. First of all, I think part of my problem is that when papers discuss these things, they claim that if $R$ is $p$-complete, sending $x\mapsto p^nx$ is topologically nilpotent on $G(R)$, in particular, it is nilpotent if $p^nR=0$. I don't understand why this is true and I would appreciate an explanation or a reference.

But if we accept the claim it is easy to see that for any connected $R$ such that $p^nR=0$, we have: $$\tilde{G}(R):=\lim_{p:G\to G}(G(R))=\operatorname{Hom}(\mathbb{Q}_p/\mathbb{Z}_p,G(R))[1/p].$$ On the other hand one has $$T_G(R):=\lim(G[n](R))=\operatorname{Hom}(\mathbb{Q}_p/\mathbb{Z}_p,G(R)).$$ So it seems to me that if $p^nR=0$, then $\tilde{G}(R)$ is just the rational Tate module, and they are different only in mixed characteristic. Is my understanding correct?