The rational approximations of $\sqrt 2$ given by its continuous fraction are: 1.5, 1.333, 1.4, 1.417, 1.412, etc. which is not strictly increasing. Similarly, this sequence for $\phi=(1+\sqrt5)\big/2$ is 2.0, 1.5, 1.667, 1.6, 1.625, etc.
Is there a real number so that the sequence of the best rational approximations (in the sense of continuous fractions) is strictly increasing?
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Recall that the $n$th approximation for $\alpha$ is $\lfloor \alpha\rfloor$ plus the $(n-1)$st approximation for $\frac1{\alpha-\lfloor \alpha\rfloor}$. You can show by induction that it does not depend on the irrational $\alpha$, but only on the parity of $n$ whether the $n$th approximation is two small or too big.
Look at the paragraph Some Useful Theorems in the Wikipedia article on continued fractions. There you will find, for instance, the statement $$\frac{h_n}{k_n}-\frac{h_{n-1}}{k_{n-1}}={(-1)^{n+1}}{k_nk_{n-1}}$$ where $\frac{h_n}{k_n}$ is the $n$th convergent of the continued fraction.
What this means is that the successive continued-fraction approximations to a positive irrational number will successively over- and under-estimate the limit. So not even three successive terms can be increasing.