This is an exam problem, so it should be rather easy but I still can't quite see how to solve it:
Let $X_1, X_2, \dots$ be independent and identically distributed random variables on the same probability space such that $$ \mathbb{P} \left[ X_1 = - \frac14 \right] = \mathbb{P} \left[ X_1 = \frac14 \right] = \frac12.$$ For all $n \in \mathbb{N}$ define $$ M_n := \prod_{k=1}^n (1+X_k).$$ Use the law of large numbers to show that $(M_n)_{n \in \mathbb{N}}$ converges almost surely and determine the limit.
(It might be worth noting that the first part was to show that $(M_n)_n$ is a martingale.)
The naive way to apply the SLLN is by taking $\log$ and then naively using SLLN. However, I can only get the limit of $\prod_{k=1}^n (1+X_k)^{\frac{1}{n}}$ in this way. So what's the correct way of applying the law of large numbers?
As James Tang pointed out $$ \mathbb{E}[\log(1+X_k)]=\frac12\log\Big(\frac{15}{16}\big)<0 $$
By SLN, $$ \frac{1}{n}\sum^n_{k=1}\log(1+X_k)\xrightarrow{n\rightarrow\infty}\frac{1}{2}\log\Big(\frac{15}{16}\Big) $$ almost surely (and in $L_1(\mathbb{P})$. A simple argument the shows that for $\mathbb{P}$--a.s all points $\omega\in\Omega$, there us $N=N(\omega)$ such that for all $n\geq N$ $$ \sum^n_{k=1}\log\big(1+X_k(\omega)\big)\leq \frac{1}{4}\log\Big(\frac{15}{16}\Big)n\xrightarrow{n\rightarrow\infty}-\infty $$ which means that $\prod^\infty_{n=1}(1+X_n)=0$ $\mathbb{P}$--a.s.