# Sequence with positive limit

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?

• 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 $\ln(1-L_k^{-1/16})$.) – 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.