Consider the symmetric random walk $S_n$ on $\mathbb{Z}$.
The process $Z_n=\exp(uS_n-n \ \log(\cosh(u)))$ for $u\in \mathbb{R}$ is a positive martingale with $E(Z_n)=1$ for all $n\geq 1$.
$Z_n$ is $L^1$ bounded and with Doob's martingale convergence theorem we get that $Z_n$ converges a.s. to an integrable random variable $Z_\infty$.
How can we prove that $Z_\infty=0$ a.s. for $u\not=0$ which would mean that $Z_n$ is not $L^1$ convergent?
I tried using the Borel Cantelli lemma but I am not sure if that is the right way.
I also tried the direct method and compute $E(\lim_{n\to\infty} Z_n)$ but I am stuck at $E(\lim_{n\to\infty} \Pi_{k=1}^n \exp(uX_k))\ \lim_{n\to\infty} \cosh(u)^{-n}$ and used Fatou's lemma to get that the expectation is less or equal than $\lim_{n\to\infty} \cosh(u)^n$. That only gives that $E(Z_\infty)\leq 1$.
Hint: Consider $n^{-1}\log Z_n$ as $n\to\infty$.