Prove that if $X$ is subgaussian, then ${\bf E}e^{tX}=1+\sum_{k=1}^{\infty}\frac{t^k}{k!}{\bf E}X^k$ 
Prove that if $X$ is subgaussian, then
  $${\bf E}e^{tX}=1+\sum_{k=1}^{\infty}\frac{t^k}{k!}{\bf E}X^k$$

So basically I just need to push the integral through the infinite sum
$${\bf E}e^{tX}=\int_{\bf R}e^{tx}d\mu_X=\int_{\bf R}\sum_{k=0}^{\infty}\frac{(tx)^k}{k!}d\mu_X$$
Thus I'll use the dominated convergence theorem, bounding the absolute value of the partial sums by the (hopefully) $\mu_X$-integrable function $e^{|tx|}$,
$$\Big|\sum_{k=0}^n\frac{(tx)^k}{k!}\Big|\leq e^{|tx|}$$
Now to show $e^{|tx|}$ is $\mu_X$-integrable, I have
$$\int_{\bf R}e^{|tx|}d\mu_X=\;\;?$$
So by the subgaussian property of $X$ I have that
$$P(|X|\geq\lambda)\leq\int_{\lambda}^{\infty}2cCxe^{-cx^2}dx=Ce^{-c\lambda^2}$$
for $c,C>0$ fixed and for any $\lambda>0$.  
Hence this integrand almost functions as my pdf for $X$, and if it did I could use the Radon-Nikodym theorem to solve this.  However even though any upper-tailed integral of it bounds that of the actual pdf, I can't quite see how to use it to bound the integral of $e^{|tx|}$.
 A: Until $\int_{\bf R}e^{|tx|}\rm d\mu_X$  things are fine (decomposition afterward needs symmetric measure). Now w.l.o.g. assume that $t>0$ and write:
$$
\int_{\bf R}e^{t|x|}\rm d\mu_X=\int_0^{\infty}e^{tx}\rm d\mu_X+\int_{-\infty}^0e^{-tx}\rm d\mu_X.
$$
We bound the first term and the second term is bounded similarly.
$$
\int_0^{\infty}e^{tx}\rm d\mu_X=\int_0^{\infty}\int_{-\infty}^{tx}e^u\rm \;du\rm\; d\mu_X\\
=\int_{-\infty}^0\int_{0}^\infty e^u\rm \;du\rm\; d\mu_X+\int_0^{\infty}e^u\int_{{u}/{t}}^{\infty}\; \rm d\mu_X \rm \;du\\
=\mu_X([0,\infty))+\int_0^{\infty}e^u
\mathbb P(X>{\frac{u}{t}})\rm \;du.
$$
Now the last integration can be bounded as:
$$
\int_0^{\infty}e^u
\mathbb P(X>{\frac{u}{t}})\rm du\leq \int_0^{\infty}e^uCe^{-cu^2/t^2}\rm\;du,
$$
which is bounded for $t>0$. Similarly one can show that $\int_{-\infty}^0e^{-tx}\rm d\mu_X$ is bounded and hence the total expectation is bounded.
A: Observe that the subgaussian property can be written as
$$P(e^{|tX|}\geq\lambda)=P(|X|\geq\frac{\log\lambda}{|t|})\leq Ce^{-c(\frac{\log\lambda}{|t|})^2}$$
Thus we have
\begin{align}
\int_{\bf R}e^{|tx|}d\mu_X={\bf E}e^{|tX|}&=\int_0^{\infty}P(e^{|tx|}\geq\lambda)d\lambda\\
&\leq C\int_0^{\infty}e^{-c(\frac{\log\lambda}{|t|})^2}d\lambda\\
&=|t|C\int_{-\infty}^{\infty}e^{-cu^2+|t|u}du\\
&=|t|Ce^{\frac{-t^2}{4c}}\int_{-\infty}^{\infty}e^{-c(u-\frac{|t|}{2c})^2}du
\end{align}
The last integral being clearly finite by reference to the Gaussian distribution.
