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I have troubles with the following problem. Can you help me, please?

Suppose $\lambda_1 = 2$ and $\lambda_2 = \frac{1}{3}$ and $\lambda_1$, $\lambda_2$ are chosen independently with probability $1/2$ each and multiply the number 1 i.e. we have a random sequence $\lambda_{i_1}1, \lambda_{i_2}\lambda_{i_1}1, \dots, \lambda_{i_n}\dots\lambda_{i_1}1$ where each $\lambda_{i_k} \in \{ \lambda_{1}, \lambda_{1} \}$.

How to show that with probability $1$, $\lambda_{i_n}\dots\lambda_{i_1}1 \rightarrow 0$?

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You could use a logarithm to turn the product into a sum. It will prove, that $$P(\lim_{n\to\infty}\sqrt[n]{1\lambda_{i_1}...\lambda_{i_n}} = \tfrac{2}{3}) = 1$$ if i'm not mistaking...

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