How are stronger $H^n \subseteq P$ or $H \subseteq \sqrt{P}$? Let $L$ be a Lie algebra over k and $H, P$ are ideals of Lie algebra $L$
Set $H^1:=H$. Define recursively $H^{n+1}:=[H,H^n]$ for all $n≥1$. Then $H^{n+1}=[H,H^n]⊆H^n$ for all $n$.
$\sqrt{P}= \cap \{M \subseteq L$ : M is prime ideal containg $P \} $
How are stronger $H^n \subseteq P$ or $H \subseteq \sqrt{P}$ ?
In other words which one implies the other 
$H^n \subseteq P \textit{ for some } n \in N $ $\implies$ $H \subseteq \sqrt{P} $ 
or 
$H \subseteq \sqrt{P}$ $\implies$ $H^n \subseteq P \textit{ for some } n \in N $
or
they are not comparable.
Any help would be appreciated.

New Update: In the case of rings.

Let $R$ be a ring, and suppose that $H, P$ are two ideals of the ring $R$ 
$$\textbf{Definition i)} \quad \sqrt{P} = \{r \in R: r^{n} \in P \textit{ for some } n \in N\}$$
$$\textbf{Definition ii)} \quad H^{n}=\left\{h^{n}: h \in H\right\} \subseteq P \textit{ for some } n \in N$$
$$1) \quad H^n \subseteq P \textit{ for some } n \in N$$
$$2) \quad H \subseteq \sqrt{P}$$ 
Proof:-
$1) \to 2$) Suppose that $H^n \subseteq P$, let $h \in H$,  then $h^n \in H^n$,  hence $h^n \in P$, because $H^n \subseteq P$, thus $h \in \sqrt{P}$. Therefore $H \subseteq \sqrt{P}$
$2) \to 1$) Suppose that $H \subseteq \sqrt{P}$, let $h^n \in H^n$,  then $h \in H$,  hence $h \in \sqrt{P}$, because $H \subseteq \sqrt{P}$, thus $h^n \in P$. Therefore $H^n \subseteq P $
 A: This is not a complete answer but I hope the ideas are helpful.
The first implication is true, that is $H^n \subseteq P \implies H \subseteq \sqrt P$. Hint: If $M$ is prime ideal containing $P$, show that $H^n \subseteq M \implies H \subseteq M$. 
I suspect the converse is not true however I do not have any immediate counterexamples.
Your proof for $(2) \implies (1)$ in the ring case is not correct. Remember that $h \in \sqrt{P}$ does not imply that $h^n \in P$ for all $n$. Unless you have some added assumptions about your ring, you cannot immediately conclude that there is a uniform constant $n$ such that $h^n \in P$ for all $h \in \sqrt{P}$. Below is an interesting example of this.
Example (Zassenhaus). Let $F$ be a field and $I = (0,1)$ be the open interval in $\mathbb{R}$. Define a vector space $R$ over $F$ with basis $\{x_i : i \in I\}$. Define multiplication on this basis by
$$
x_i x_j = \left\{
\begin{array}{!}
x_{i+j} & \textrm{if } i+j < 1, \\
0 & \textrm{otherwise.}
\end{array}
\right.
$$
Extend this multiplication linearly to $R$. Let $P$ be the ideal generated by $\{x_i : i > 1/2 \}$. If you play a little with this ring you'll see that $\sqrt P = R$. However for all $n > 0$, we have $(x_{1/(2n+1)})^n \not\in P$
A: $$H \subseteq \sqrt{K}  \implies H^n \subseteq K \textit{ for some    } n \in N $$
Proof by contrapositive:- (P $\implies$ Q) $\equiv $ ( ¬Q $\implies$ ¬P)
Let $H$ and $K$ be two ideals of Lie algebra $L$ such that $H^n \not  \subseteq K $
Now, suppose that $x \in K-H^n$ for some $n \in N$
$\implies$ $x \in K \subseteq \sqrt{K}  $, and $ x \not \in H^n$ for any $n \in N$ thus $x \not \in H^1=H.$
$\implies$ $x \in \sqrt{K}-H $
$\implies$ $H \not \subseteq \sqrt{K}$
Therefore $H \subseteq \sqrt{K}  \implies H^n \subseteq K \textit{ for some } n \in N $
A: $$H \subseteq \sqrt{K}  \implies H^n \subseteq K \textit{ for some    } n \in N $$
Proof by contrapositive:- (P $\implies$ Q) $\equiv $ ( ¬Q $\implies$ ¬P)
Let $H$ and $K$ be two ideals of Lie algebra $L$ such that $H^n \not  \subseteq K $
Now, suppose that $x \in H^n - K$ for some $n \in N$
$\implies$ $x \in H^n \subseteq H  $ and $ x \not \in K$ for any $n \in N$.
$\implies$ $x \in H-K \subseteq H-\sqrt{K} $
$\implies$ $H \not \subseteq \sqrt{K}$
Therefore $H \subseteq \sqrt{K}  \implies H^n \subseteq K \textit{ for some } n \in N $
