I posted this question a few days ago and there were some errors in my post. I have fixed them and it should be all right now. Hope someone can help with my confusion.
Let $(X_n)_{n\ge0}$ denote an infinite sequence of random variables such that ${X_n} = (1 + \varepsilon )^n{X_0}$ where $\varepsilon$ is a small positive number. Clearly ${X_n} \to \infty $ as $n \to \infty$.
Now the question is
Given $0< 2\varepsilon < \sigma^2 < 1$ and some i.i.d noise $(\xi_i)$ satisfying $\Bbb E \xi=0$, $\Bbb E \xi^2=1$, $\Bbb E \xi^3<\infty$ and $|\varepsilon + \sigma\xi| < 1$, define ${X_n} = X_0\prod\limits_{i = 1}^n ({1 + \varepsilon + \sigma {\xi _i}})$. Show that ${X_n} \to 0$ as $n \to \infty$ almost surely.
Attempt: $$\log |{X_n}| = \log |\prod\limits_{i = 1}^n {(1 + \varepsilon + {\sigma\xi _i})} ||{X_0}| = n(\frac{1}{n}\sum\limits_{i = 1}^n {\log (1 + \varepsilon + {\sigma\xi _i})} ) + \log |{X_0}|$$
By strong law of large numbers $ \frac{1}{n}\sum\limits_{i = 1}^n {\log (1 + \varepsilon + {\sigma\xi _i})} \to \Bbb E {\log (1 + \varepsilon + {\sigma\xi})}$ as $n \to \infty$ almost surely. So if we can prove $\Bbb E {\log (1 + \varepsilon + {\sigma\xi})} < 0$, then it leads to $\log|X_n| \to -\infty$ as $n\to\infty$, and so $X_n \to 0$.
By Taylor expansion (since we have $|\varepsilon + \sigma\xi| < 1$), we can get something like $$\log (1 + \varepsilon + \sigma \xi ) = (\varepsilon + \sigma \xi ) - \frac{1}{2}{(\varepsilon + \sigma \xi )^2} + o{(\varepsilon + \sigma \xi )^2}$$
I can show that for the first two terms of the expansion $$\Bbb E[(\varepsilon + \sigma \xi ) - \frac{1}{2}{(\varepsilon + \sigma \xi )^2}] = \varepsilon - \frac{1}{2}{\varepsilon ^2} - \frac{1}{2}{\sigma ^2} < 0$$ since $0< 2\varepsilon < \sigma^2 < 1$.
But I got stuck when I try to prove the expectation is negative in general, that is to show $$\Bbb E\log (1 + \varepsilon + \sigma \xi ) = \Bbb E [(\varepsilon + \sigma \xi ) - \frac{1}{2}{(\varepsilon + \sigma \xi )^2} + o{(\varepsilon + \sigma \xi )^2}] = \varepsilon - \frac{1}{2}{\varepsilon ^2} - \frac{1}{2}{\sigma ^2} + \Bbb E o{(\varepsilon + \sigma \xi )^2} < 0$$
I am confused on how to deal with the infinitesimal when it is under expectation.
Thank you!