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Let $b\in[1,e]$ and define $$ a_{n+1}=\left(\sqrt[b]{b}\right)^{a_n} $$ with $a_0=\sqrt[b]{b}$. Show that the sequence is convergent and find the limit.


One can show that $(a_n)$ is increasing. To show it is convergent, it suffices to give a bound for $(a_n)$. I get stuck here and I don't see how to go on. Some manipulation with the formula give $$ a_{n+1}=\exp\left(a_n\log\sqrt[b]{b}\right)=\exp\left(a_n\log\left(e^{\frac{1}{b}\log b}\right)\right)=\exp\left(a_n\frac{\log b}{b}\right), $$ which doesn't seem to be of much help.

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  • $\begingroup$ I obtained that limit is equal to $b$, but can't prove it without a pen. $\endgroup$ – user98186 Mar 1 '16 at 21:26
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You're almost there. Note that $(\log b)/b \leq 1/e$ so $\log a_0 \leq 1/e$ and $a_{n+1} \leq \exp(a_n/e)$. Then use induction.

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