Find the moment generating function (MGF) of $$\frac{U_1 + U_2 + \dots + U_n}{\sqrt{n}}$$ where $U_i \sim \mbox{Unif}(-1,1)$.
I know that the MGF of a sum of independent random variables is equal to the product of the MGF of each random variable, so that in this case $$M(U_1+U_2+\dots+U_n)= \left(E\left(e^{tU_i}\right)\right)^n = \frac{e^t - e^{-t}}{2t}$$ but what do you do about the $\sqrt{n}$ in the denominator? Are you allowed to say that $$M\left(\frac{U_1 +\dots+U_n}{\sqrt{n}}\right)=\frac{\left(E\left(e^{tU_i}\right)\right)^n}{\sqrt{n}}\text?$$
Let $S_n = \sum_{k=1}^n U_k$ where $U_k \sim \mathcal{U}(-1,1)$.
It's not clear from your question if you mean discrete uniform or continuous uniform distribution? You seems to be using the MGF from the continuous one, not sure if it is intended, assuming so.
The MGF indeed is $$M_U(t) = \frac{e^t-e^{-t}}{2t}$$ and for the combined one it is $$ M_S(t) = \prod_{k=1}^n M_U(t) = M_U(t)^n= \left( \frac{e^t-e^{-t}}{2t} \right)^n. $$ Finally, there is a question of scaling. For $a\in \mathbb{R}^+$, let $X = S/a$, then $$ M_X(t) = \mathbb{E}\left[e^{tX}\right] = \mathbb{E}\left[e^{St/a}\right] = M_S(t/a). $$ Can you now plug that back into $M_S(t)$ we used above, letting $a = \sqrt{n}$ and examine what happens when $n \to \infty$?