Relation between Jacobson radical and composition series 
Let $R$ be a not necessarily commutative ring with 1. Suppose $R$, viewed as a right $R$-module, has a finite composition series with non-isomorphic composition factors. Prove that the Jacobson radical of $R$ is equal to $0$.

I can't see the relation between Jacobson radical and composition series at this question. 
Please give me clue about what it can be, i think question will be easy then.
 A: If $S$ is a simple $R$-module, then $S \cong R/M$ for a maximal (one-sided) ideal $M$. Hence every simple module $S$ is isomorphic to a quotient module of $R/J(R)$ for $J(R)$ the Jacobson radical (which is the intersection of the maximal (one-sided) ideals $M$).
In particular, if $R$ is a nice ring where $R/J(R)$ has a composition series, then the composition factors include every simple module $S$ (by Jordan–Hölder). Some of those simple modules could be repeated. Indeed, Artin-Wedderburn: if $R/J(R)$ has a composition series, then $R/J(R)$ is $Z(R)$-algebra isomorphic to a direct sum of matrix rings over division $Z(R)$-algebras, each matrix ring corresponding to exactly one isomorphism class of simple module, and the dimension of the matrix ring over the division algebra is how many times that simple module appears in a composition series of $R/J(R)$.
If $R$ itself has a composition series, then so does $R/J(R)$ and $J(R)$, and so all of the composition factors of $J(R)$ are actually repeats of composition factors of $R/J(R)$.
So if $R$ has a composition series with no repeated composition factors then $J(R)=0$ (it has no composition factors left) and $R/J(R)$ is a direct sum of finitely many division $Z(R)$-algebras. In other words, $R$ is a direct sum of finitely many fields and/or skew fields.
A: It seems that we can borrow the idea from How to prove that $\mathrm{End}_R(M)$ is a division ring.
Following the comment of Sammy Black on Feb 13, 2014 (but here we write the composition series downwards as $$R_R =: M_0 \supset M_1 \supset \cdots \supset M_n = 0,$$
we see that $M_i J(R) \subset M_{i+1}$ for all $i=0, 1, \ldots, n-1$. Then we start from $M_0 = R_R$ inductively, obtaining
$$
J(R)^n = 0.
$$
So actually we have shown the following result: if the regular module $R$ is of finite length, then its Jacobson radical $J(R)$ is nilpotent. Note that $J(R)$ is not always nilpotent in general.

Anyway, one knows that if $R$ is left artinian, then $J(R)$ is already nilpotent. (See Lam A First Course in Noncommutative Rings, (4.12).) So the argument above by composition series proved something weaker. If one knows the result above, the arguments we carried out above is useless.

Now the extra condition that $R_R$ has nonisomorphic composition factors enables us to conclude that $J(R)=0$.
To show $J(R)=0$, suppose otherwise we have an element $u \neq 0$ in the Jacobson radical. Consider the endomorphism of $M$ by right multiplication of $u$, and denote it as $u$ by abuse of notation. Then we see that $u$ is a nilpotent endomorphism.
Now consider the short exact sequence
$$
0 \rightarrow \ker(u) \rightarrow R_R \rightarrow u(R_R) \rightarrow 0.
$$
Then we can derive the information of composition factors from this:

*

*$\mathrm{JH}(R_R) = \mathrm{JH}(\ker(u)) \cup \mathrm{JH}(u(R_R)).$

*$\mathrm{JH}(\ker(u)) \cap \mathrm{JH}(u(R_R)) = \emptyset$ since the composition factors of $R_R$ are nonisomorphic.

This implies the restriction $u|_{u(R_R)}: u(R_R) \rightarrow u^{2}(R_R)$ is injective.(For more on this, see the link at the beginining of this answer.) Now since $u \neq 0$, there exists $x_0 \in R$ such that $u(x_0) \neq 0$. (In fact, taking $x_0 = 1_R$ will do.) Then repeatedly, we have a sequence of nonzero elements $u^n(x_0) \in R$, contradicts to that $u$ is a nilpotent element in $J(R)$, hence in $R$.
Hope that I got this right. :)
