The limit of an expected value vs expected value of a limit in this betting game Setting
The outcome $X$ of a slot machine takes values 1,2,or 3 with probability $p(1) = \frac{1}{2}$, $p(2) = \frac{1}{4}$, $p(3) = \frac{1}{4}$. We are given 3 for one odds, that is if we bet 1 dollar and win then we will win $3. 
The betting strategy is as follows: we bet $\frac{1}{2}$ of current wealth that $X = 1$, $\frac{1}{4}$ of wealth that $X = 2$, and $\frac{1}{4}$ of wealth that $X = 3$.
So our winning after n rounds is
$$V_n = 3 V_{n-1} p(X_n).$$
Now find $\lim_n V_{n}$. 
Attempted Solution
So I take the expected value of $V_n$ for all n and then take the expected value to the limit:
$$E[V_n] = \sum_{k=0}^n \binom{n}{k} \left(\frac{3}{4}\frac{1}{2}\right)^{n-k}\left(\frac{3}{2}\frac{1}{2}\right)^{k} = \left(\frac{9}{8}\right)^n$$
And $\left(\frac{9}{8}\right)^n \rightarrow^n \infty$.
Confusion
However, the counter argument to the solution is that we have taken then limit of an expectation, when in fact, I need to the expectation of the limit. And by Fatou's lemma
$$\lim inf E[V_n] \ge E[\lim inf V_n]$$
So I found an upper bound that diverges and therefore cannot conclude anything about $V_n$. Please explain the subtleties here?
Alternate Solution
Now consider this alternate solution and please explain why it makes sense relative to Fatou's lemma. Define $\log := \log_2$. And note
$$V_n = 3^n V_o  P(X_1) \cdot \ldots \cdot P(X_n)$$
So let $V_o = 1$ and we have: 
$$2^{\log V_n} = 2^{n \log 3 + \log V_o + \log P(X_1) + \ldots + \log P(X_n)}$$
And by LLN $\log P(X_1) + \ldots + \log P(X_n) \rightarrow^n n E[\log(X_j)] = n \left(\frac{1}{2} \log \frac{1}{2} + \frac{1}{2} \log \frac{1}{4} \right)$.
And the rest is just algebra. 
But my primary concern is how does taking a limiting argument for the sum of $P(X_j)$'s sit wrt Fatou's lemma. 
 A: Expected values can be misleading.  We have
$$ \mathbb E[V_{n+1}|V_{n}] = \dfrac{1}{2} \cdot \dfrac{3}{2} V_n + \dfrac{1}{2}\dfrac{3}{4} V_n = \dfrac{9}{8} V_n $$
so that $\mathbb E[V_n] = (9/8)^n V_0 \to \infty$.  
However, the Strong Law of Large Numbers is about sums, not products.
To make a product into a sum, we take logarithms.
$$ V_n = V_0 \prod_{j=1}^n X_j \ \text{so} \log V_n = \log V_0 + \sum_{j=1}^n \log X_j $$
with
$$\mathbb E[\log X_j] = \dfrac{1}{2} \log (3/2) + \dfrac{1}{2} \log(3/4) = \dfrac{1}{2} \log(9/8)$$
This is still positive, so it's still a good bet, but not quite as good
as what the expected value would lead you to think.  What the Strong Law
says is that almost surely, $$\dfrac{1}{n} \log V_n \to \dfrac{1}{2} \log(9/8)
\ \text{as}\ n \to \infty$$
So for any $\epsilon \in (0,9/8)$, with probability $1$ we can say that for large enough $n$, $(9/8- \epsilon)^{n/2} V_0 < V_n < (9/8 + \epsilon)^{n/2} V_0$.  In particular, $\lim_{n \to \infty} V_n = +\infty$.  But notice that
this almost-sure growth of $V_n$ is much slower than the growth of the expected value.
For an example where the expected value grows but the investment almost surely goes to $0$, see this Maple application that I wrote a while ago.
A: Each round, you have $1/2$ chance of multiplying your money  by $3/2$ and $1/2$ chance of multiplying it by $3/4$.  You are correct that the expectation after one round is $9/8$. For any finite number of rounds, your expectation is then $(9/8)^n$.  After $n$ rounds you have a distribution of results.  If you have won $m$ times, you have $\frac {3^n}{2^n \cdot 2^{n-m}}$. For finite $n$ you can calculate the expected value over this distribution.  The distriution does not have a limit as $n \to \infty,$ so you can't calculate its expectation.
