<s>[INST] 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.
 [/INST] 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}
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