Suppose $E$ is a spectrum and $p$ is a prime. We can then $(H\mathbb{Z}/p)\_\*$-localize to obtain $L\_{H\mathbb{Z}/p}E$. Is it true that the natural map $L\_{H\mathbb{Z}/p}E\rightarrow\text{holim}\_n(L\_{H\mathbb{Z}/p}E)/p^n$ is an $(H\mathbb{Z}/p)\_\*$-equivalence?



Homotopy limit and Bousfield localization
Yes, $H\mathbb{F}\_p$-localizations are $p$-complete. We have the cofiber sequence
$$
\Sigma^{-1}\mathbb{S}/p^\infty \xrightarrow{j} \mathbb{S}\to \mathbb{S}[p^{-1}] \to \mathbb{S}/p^\infty
$$
where $\mathbb{S}/p^\infty\approx \mathrm{colim}\_n \mathbb{S}/p^n$, from which it is not hard to see that the $p$-completion map $X\to \mathrm{holim}\_n X/p^n=X^\wedge\_p$ is equivalent to 
$$
F(\mathbb{S},X) \xrightarrow{j^\*} F(\Sigma^{-1}\mathbb{S}/p^\infty, X).
$$
In particular, $X\to X^\wedge\_p$ is an equivalence iff $F(\mathbb{S}[p^{-1}], X)=0$.


We have that $H\mathbb{F}\_p\wedge \mathbb{S}[p^{-1}]=0$, so $F(\mathbb{S}[p^{-1}], L\_{H\mathbb{F}\_p}X)=0$.


