Prove that the sum of all the nil right ideals of $R$ is a 2-sided ideal Prove that the sum of all the nil right ideals of $R$ is a nil 2-sided ideal.
I need help showing that that the sum of all nil right ideals is is a nil ideal. I've shown it is an ideal already:
$proof:$
First note that if $I$ is a right nil ideal of $R$ and $r \in R$ then $rI$ is also a right nil ideal. This is because of $(xr)^n=0$ then $(rx)^{n+1}=0$
Let $S$ be the sum of all nil right ideals of R and let $y \in S$.
Then $y = y_1 + y_2 + ... +y_k$ where $y_i \in I_i$ where $I_i$ is a right ideal of $R$. 
Then $ry = r(y_1 + y_2 + ... +y_k)$ and $ry_i \in rI_i$. 
And $S$ is obviously a right ideal.
Thus $S$ is an ideal of $R$. Showing that it is nil is proving to be a little harder, maybe I'll need a generalized binomial formula? Can somebody help walk me through this?
Thanks!
 A: I take it from your mention of "right ideal" in your work that you're dealing with noncommutative rings.
A proof of your question would amount to a proof of the Köthe conjecture which is still unsolved.  (If the sum of all nil right ideals is guaranteed to be a nil ideal, then in particular the sum of any two nil right ideals is a nil right ideal.)
I notice that the title of your question simply says "Prove that the sum of all the nil right ideals of  is a 2-sided ideal" which is true, as you demonstrated. Is it possible that your idea to prove it is additionally nil something that just slipped in later by accident?
A: Note: This answer only shows that the sum of nil ideals is nil in a commutative ring.
As pointed out by rschwieb, that question is open for non-commutative rings. I'll leave this here in case anyone finds the argument helpful in proving the commutative case.

We can avoid having to use the multinomial theorem, which would be messy. Elements $s\in S$ are of the form $s=x_1+\cdots+x_n$ for some $n\geq1$ and $x_i\in I_i$, where $I_i$ is a nil ideal. Then we can use induction on $n$ to show that each $s$ is nilpotent.
The result is obvious for $n=1$. If we have $s=x_1+\cdots+x_n$, then $r=x_1+\cdots+x_{n-1}$ is nilpotent by the inductive hypothesis, say $r^a=0$ for some $a\geq1$. We know that $x_n^b=0$ for some $b\geq1$. Then by the usual binomial theorem we have $$s^{a+b}=(r+x_n)^{a+b}=\sum_{k=0}^{a+b}\binom{a+b}{k}r^{a+b-k}x_n^k$$ From this, we can see in the sum that as soon as we have fewer than $a$ lots of $r$ we will have at least $b$ lots of $x_n$, and so the sum will be $0$. Then $s^{a+b}=0$, so $s$ is nilpotent and we are done by induction.
