about an interesting moment generating function Let $X_1,\ldots,X_n$ be iid Rademacher variables, i.e., $P(X_1=1)=P(X_1=-1)=1/2$. CLT says that $Y_n\equiv \sqrt{n}\bar{X}$ converges in distribution to $N(0,1)$ as $n\to\infty$. So $Y_n^2$ is asymptotically $\chi^2$ which means $E e^{tY_n^2}=O(1)$ provided that $t<1/2$. Intuitively, $E e^{t Y_n^3}\to\infty$ since the third power of standard normal variable has diverging moment generating function.
My question is what can we say about $E e^{t Y_n^3/\sqrt{n}}$? Intuitively $Y_n/\sqrt{n}\to0$ almost surely so the expectation seems to be bounded by $E e^{o(1) Y_n^2}$, which indicates that $E e^{t Y_n^3/\sqrt{n}}=O(1)$ for any $t>0$. But I don't know how to show this rigorously. 
 A: Let us prove 

Theorem 1: There is some $t_*\in[0.672,0.694]$ such that $E_n:=E e^{t Y_n^3/\sqrt{n}}$ is bounded for $t\in(0,t_*)$ and unbounded for $t\in(t_*,\infty)$. 

Proof. The unboundedness of $E_n$ for $t\ge\ln2=0.693\ldots$ follows by Christian Remling's comment. It remains to show that $E_n$ is bounded for $t\in(0,0.672]$. By the Chebyshev--Bernstein inequality with $z=\sqrt n\,\tanh^{-1} y/\sqrt n$, for each real $y\in[0,\sqrt n)$ we have 
\begin{equation}
 P(Y_n>y)\le e^{-zy}E e^{zY_n}=e^{-zy}\cosh^n\frac z{\sqrt n}=e^{-ng(u)}, 
\end{equation}
where $\tanh^{-1}$ is the function inverse to $\tanh$, $u:=y/\sqrt n$, and 
\begin{equation}
 g(u):=u \tanh^{-1} u-\frac12\,\ln\frac1{1-u^2}. 
\end{equation}
So, integrating by parts, we have 
\begin{align}
 E_n&=E e^{t Y_n^3/\sqrt{n}}1_{Y_n\le0}+E e^{t Y_n^3/\sqrt{n}}1_{Y_n>0} \\ 
 &\le P(Y_n\le0)-\int_0^\infty e^{t y^3/\sqrt{n}}\,dP(Y_n>y) \\ 
 &=P(Y_n\le0)+P(Y_n>0)+\int_0^\infty \frac{3ty^2}{\sqrt n}e^{t y^3/\sqrt{n}}\,P(Y_n>y)\,dy \\ 
 &\le1+3tn\int_0^1 u^2\exp\{n(t u^3-g(u)\}\,du. 
\end{align}
Next, $g(u)>t_1 u^3$ for $t_1:=0.67208$ and all $u\in(0,1)$. So, for $h:=t_1-0.672>0$ and $t\in(0,0.672]$, 
\begin{equation}
  E_n\le1+3tn\int_0^1 u^2\exp\{n(t-t_1)u^3\}\,du<1+\frac t{t_1-t}=\frac{t_1}{t_1-t}\le\frac{t_1}h=O(1), 
\end{equation}
as desired.
