Solution
= Solution
The two-point distribution gives $M(t)=(e^t+e^{-t})/2=\cosh t$. In the power series, $(2j)!=\prod_{r=1}^j(2r)(2r-1)\geq2^jj!$, so each coefficient satisfies
$$
\frac{t^{2j}}{(2j)!}\leq\frac{(t^2/2)^j}{j!}.
$$
Summing the nonnegative terms proves $\boxed{M(t)\leq e^{t^2/2}}$ for every real $t$. Part (a) therefore gives
$$
\boxed{M_{S_n}(t)\leq e^{nt^2/2}.}
$$
This is the <sub-Gaussian random variable>[sub-Gaussian] moment bound for a symmetric random sign.