Suppose that $E[X^n] = 3n$. Find $E[e^X]$... Suppose that $E[X^n] = 3n$. Find $E[e^X]$. 
Hint from my professor: $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} +···$
Not quite sure how to solve this problem, wouldn't $e^x$ go on exponentially. Any help is really appreciated.
 A: Warning: What appears below is at best vacuously true --- i.e. true but uninformatitve, since, as several people have pointed out, there is not actually any probability distribution whose $n$th moment, for every positive integer $n$, is $3n$.
\begin{align}
\operatorname{E}\left(e^X\right) & = \operatorname{E}\left( 1 + X + \frac{X^2} 2 + \frac{X^3} 6 + \frac{X^4}{24} + \cdots \right) \\[10pt]
& = 1 + \operatorname{E}(X) + \frac{\operatorname{E}(X^2)} 2 + \frac{\operatorname{E}(X^3)} 6 + \frac{\operatorname{E}(X^4)}{24} + \cdots \\[10pt]
& = 1 + 3 + \frac 6 2 + \frac 9 6 + \frac{12}{24} + \cdots+ \frac{3n}{n!} + \cdots \\[10pt]
& = 1 + \frac 3{0!} + \frac 3 {1!} + \frac 3 {2!} \cdots + \frac 3 {(n-1)!} + \cdots \\[10pt]
& = 1 + 3\left( \frac 1 {0!} + \frac 1 {1!} + \frac 1 {2!} + \frac 1 {3!} + \cdots \right) \\[10pt]
& = 1 + 3e.
\end{align}
A: We show that there is no such random variable. 
Note: we assume  $E [X^n] = 3n$ for $n\ge 1$ (because, as commented above,  the case $n=0$ cannot hold).
Recall that for real-valued random variable $Y$: 
$$E( Y^2) \ge (E Y)^2$$ 
provided the second moment is finite (variance $E [Y^2- (E Y)^2] = E(Y^2)-(EY)^2$ is nonnegative). In particular, letting $Y=X^n$ for $n\ge 1$.  
$$ 6n = E [X^{2n}] \ge (E [X^n])^2= (3n)^2,$$ 
or $n <6/9 <1$, so no such random variable exists !
