Let $P$ be a prime ideal in a commutative ring $R$ with unity such that an ideal $Q$ is $P$-primary and some power of $P$ is a subset of $Q$. I want to show that $\sqrt {Q[[x]]}=P[[x]]$.
If a power series $f=a_0+a_1x+a_2x^2+\cdots$ is in $\sqrt {Q[[x]]}$ then some power $f^k$ is in $Q[[x]]$ so that $a_0^k$ belongs to $Q$ whence, by the hypothesis, $a_0$ would be in $P$. But for the other coefficients $a_i$ I could not reach the same result. Thanks for any help!
From $Q\subseteq P$ we get $Q[[X]]\subseteq P[[X]]$, and therefore $\sqrt{Q[[X]]}\subseteq\sqrt{P[[X]]}=P[[X]]$ (why?).
Now if we want to show $P[[X]]\subseteq\sqrt{Q[[X]]}$ start with a power series $f\in P[[X]]$, say $f=a_0+a_1X+\cdots$, and consider the ideal $I=(a_0,a_1,\dots)$. Since $P^k\subseteq Q$ we also have $I^k\subset Q$, and therefore $f^k\in Q[[X]]$, that is, $f\in\sqrt{Q[[X]]}$.