# Sum of all solvable ideals of a Lie algebra and radical

Let $$\mathfrak{g}$$ be a finite dimensional Lie algebra. I know the fact that if the ideals $$\mathfrak{a}$$,$$\mathfrak{b}$$ are solvable, then so is $$\mathfrak{a+b}$$.

Now I want to show the existence of maximal solvable ideal (called "radical") of $$\mathfrak{g}$$ by showing that the (infinite) sum over all solvable ideals in $$\mathfrak{g}$$ is solvable. But why is this infinite sum of solvables again solvable(does the above fact apply immediately?)? Or should I prove the existence of radical in another way?

If $$dim{\cal G}=n$$ is finite. Let $$A_1, A_2$$ two non zero solvable ideals, suppose that $$A_2$$ is not contained in $$A_1$$, $$dim(A_1+A_2)>dim(A_1)>2$$. Suppose that there exists $$A_3$$ solvable not contained in $$A_1+A_2$$, $$dim(A_1+A_2+A_3)>dim(A_1+A_2)>3$$,... The sequence needs to stop we cannot construct $$A_n$$, since it will implies that $$dim(A_1+A_2+...+A_n)>n>dim({\cal G})$$.
This implies that there exists $$p$$ such that every solvable ideal $$B$$ is contained in $$R=A_1+...+A_p$$; $$R$$ is the radical.