Show that for Lebesgue-almost every x∈[0,1), the geometric mean
lim
exists and has common value. What is this? (no proof required)
I think this has something to do with the Birkhoff ergodic Theorem
I tried \begin{align} \log \left( \lim_{n \rightarrow \infty} \left(\prod_{i=1}^{n} (a_i+1) \right)^{1/n} \right) &= \lim_{n \rightarrow \infty} \log\left(\prod_{i=1}^{n} (a_i+1) \right)^{1/n} \\ &= \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{i=1}^{n} \log (a_i+1) \\ &= ....??? \end{align}
It was shown in the part before that if x = \sum_{i=1}^{\infty}\frac{a_i}{10^i} where a_i \in \{0,1,\dots,9 \} that for Lebesgue-almost every x \in [0,1) that
\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{i=1}^{n}a_i=\frac{9}{2}
but I cannot see how this can be used.
Answer
The nth term is \exp(S_n(x)/n) where S_n(x)=\sum\limits_{k=1}^nX_k(x),\qquad X_k(x)=\log(1+a_k(x)). With respect to the Lebesgue measure on [0,1), the sequence (a_k) is i.i.d. hence (X_k) is i.i.d. and S_n\to E(X_1) almost surely, by the strong law of large numbers for i.i.d. integrable sequences. Furthermore, a_1 is uniform on \{0,1,\ldots,9\} hence E(X_1)=\frac1{10}\sum\limits_{i=0}^9\log(1+i)=\frac1{10}\log(10!).
Thus, \exp(S_n(x)/n)\to\ell for almost every x, where \ell=\exp(E(X_1))=(10!)^{1/10}\approx4.5287, and in particular, \ell\ne9/2.
Nota: One may replace "the strong law of large numbers for i.i.d. integrable sequences" above by "Birkhoff ergodic theorem".
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