# Making a Logarithm grow faster than a linear function.

Suppose $\theta \in [0,1]-\mathbb{Q}$ has a sequence of principal convergents $(\frac{m_k}{n_k})$, obtained from the continued fraction representation $\theta=[0;a_1,a_2,...].$ Let $0<\varepsilon \ll \delta$. I'm wondering if $(1+\delta)\log n_k + (k+1)\log 2< (1+\delta +\varepsilon)\log n_k$ for sufficiently large $n_k$. Basically this boils down to a linear $k+1$ versus a log. My question is: is there a a constraint I can place on $\theta$ so that $\log n_k$ grows faster than $k+1$?

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The faster the partial quotients grow, the faster $n_k$ grows. – Gerry Myerson Nov 7 '12 at 6:29
Yes, the partial quotients should be growing fast, but how do I determine if something is growing so fast that the log of it is growing faster than the limit itself? – The Substitute Nov 7 '12 at 6:31
Actually, $\log n_k$ can hardly fail to grow faster than $k+1$. Even if all the partial quotients are 3, $n_k$ will exceed $3^k$, so $\log n_k$ will exceed $(\log 3)k$, which exceeds $k+1$. – Gerry Myerson Nov 7 '12 at 12:02
yes! thank you; that's great. Put that as an answer if you want me to "favorite" it. – The Substitute Nov 8 '12 at 4:06

Unless the partial quotients are mostly very small, $\log n_k$ will grow faster than $k+1$. For example, if the partial quotients are all $3$, then $n_k$ will exceed $3^k$, and $\log n_k$ will exceed $(\log 3)k$, which exceeds $k+1$.