Proving this random variable problem $X_1,X_2,X_3,\ldots$ are IID random variable taking values in $(-1,\infty)$.
Also $t\in(0,1)$.
Define random variables $Y_1,Y_2,Y_3,\ldots$ recursively like
$$Y_1 = (1+tX_1)$$
$$Y_n = Y_{n-1}(1+tX_n)$$
Imagine this is money being partially invested someplace again and again.
How can I prove that
$$Y_n\to\infty\,\, \textrm{a.s.}\,\,\,\,\textrm{if and only if}\,\,\,\, E[\log(1+tX_1)] > 0 \,\,\,\,\,\textrm{ ?}$$
If you can help me solve this, I'd also appreacite help with a similar problem but generalized: Solving this random variable problem
 A: This is not a formal proof by any means, but more like an idea of how to do it. Maybe it'll help you to get started. 

Note that you may write
$$
Y_n=\prod_{i=1}^n (1+tX_i).
$$
Furthermore, the log of this is
$$
\log Y_n=\sum_{i=1}^n \log(1+tX_i).
$$
By the law of large numbers,
$$
\frac{\log Y_n}{n}=\frac{\sum_{i=1}^n\log(1+tX_i)}{n}\longrightarrow E\left[\log(1+tX_1)\right] \quad \text{as } n\to \infty,
$$
where by the iid assumption we may use $X_1$ in the expectation.
For almost sure convergence, you want
$$
Pr\left(\lim_{n\to \infty} Y_n=\infty\right)=1.
$$
If $Y_n$ goes to infinity, then $\log Y_n$ will do so too. Hence, we can use
$$Pr\left(\lim_{n \to \infty} \log Y_n = \infty\right)=1.$$
The key is to show that if $\frac{\log Y_n}{n}$ converges to a non-zero finite constant, then this implies that $\log Y_n$ tends to infinity. 
A: Note that
$$\log(Y_n) = \sum_{i=1}^{n}\log{(1+tX_i)}> \sum_{i=1}^{n}\frac{\log(1+tX_i)}{i} $$
By the SLLN, almost surely, 
$$\frac{\log(Y_n)}{n} = \frac{1}{n}\sum_{i=1}^{n}{\log(1+tX_i)} \rightarrow E[\log(1+tX_1) > 0 \\ $$
An application of the Cesaro Mean Theorem says that if $\sum_{i=1}^{n}\frac{\log(1+tX_i)}{i}$ coverges to a finite limit, then  $\frac{1}{n}\sum_{i=1}^{n}\log(1+tX_i) \rightarrow 0$, which is not the case.
Therefore $\log(Y_n) \rightarrow \infty$ and $Y_n \rightarrow \infty$.
