Consider $L_0=100$, and $L_{k+1}=L_k^3$, for $k \in \mathbb{N}$.

Now, consider $\rho_0>0$ and $\rho_{k}=\rho_{k+1}(1-L_k^{-\frac{1}{16}})$.

The book I'm reading states that the sequence $(\rho_k)_{k \in \mathbb{N}}$ has a positive limit. Clearly this sequence is descreasing because $1-L_k^{-\frac{1}{16}}<1$. Besides that, $ \rho_k \in(0, \rho_0)$ and therefore $(\rho_k)_{k \in \mathbb{N}}$ is convergent. But I can't conclude that the limit is positive. Can anybody help me?

  • $\begingroup$ The sequence is actually increasing, not decreasing: the inequality $1-L_k^{-1/16}\lt1$ implies $\rho_k=\rho_{k+1}(1-L_k^{-1/16})\lt\rho_{k+1}$. (The accepted answer is nonetheless still essentially correct, it just needs a minus sign in front of the $\ln(1-L_k^{-1/16})$.) $\endgroup$ – Barry Cipra 2 days ago

One has $$\ln(\rho_{k+1})=\ln(\rho_{k})+\ln(1-L_k^{-1/16}).$$ And so, since $L_i=L_0^{3^i}$, we have $$\ln(\rho_{k+1})=\sum_{i=0}^k \ln (1-L_i^{-1/16}) = -\sum_{i=0}^k \sum_{j=1}^\infty \frac{1}{j}(L_i^{-1/16})^j=-\sum_{i=0}^k \sum_{j=1}^\infty \frac{1}{j}(L_0^{-3^i/16})^j.$$ If $\rho_k \rightarrow 0$ as $k\rightarrow\infty$, we must have $$\sum_{i=0}^\infty \sum_{j=1}^\infty \frac{1}{j}(L_0^{-3^i/16})^j=\infty.$$ However, since the series $\sum_{i=0}^\infty (L_0^{-j/16})^{3^i}$ is bounded by $\sum_{t=1}^\infty (L_0^{-j/16})^{t}$ which is in turn bounded by $cL_0^{-j/16}$, where $c=L_0^{1/16}$, we have $$\sum_{i=0}^\infty \sum_{j=1}^\infty \frac{1}{j}(L_0^{-3^i/16})^j=\sum_{j=1}^\infty \frac{1}{j}\sum_{i=0}^\infty (L_0^{-3^i/16})^j= \sum_{j=1}^\infty \frac{1}{j}\sum_{i=0}^\infty (L_0^{-j/16})^{3^i} \leq c\sum_{j=1}^\infty \frac{1}{j}L_0^{-j/16}.$$ But since $L_0^{-j/16} <1/j$ for $j$ large enough, the series on the right is convergent, a contradiction.


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