Let $I\_{\lambda},$ $\lambda>0$ be a subset of all irrational numbers $\rho=[a\_{1},a\_{2},...,a\_{n},...]\in(0,1)$ such that $a\_{n}\leq \text{const}\cdot n^{\lambda}.$


Here, $[a\_{1},a\_{2},...,a\_{n},...]$ is the continued fraction with partial quotients $a\_1,a\_2,\dots$. 


My question is: under what values of $\lambda$ the set $I\_{\lambda}$ has a positive Lebesgue measure? 



Lebesgue measure of a set of irrational numbers
The condition $\lambda>1$ is sufficient and, at least almost, necessary:


To clarify, the space of irrational numbers $(0,1)-\mathbb Q$ is homeomorphic to
$\omega^\omega$ under the map that sends $\frac{1}{a\_1+\frac{1}{a\_2+\cdots}}$ to a function $f$ satisfying
$f(n)=a\_{n+1}-1$. (As is well known.)


This way Lebesgue measure on $(0,1)$ induces a "Gaussian" measure on $\omega^\omega$.


Under this measure $\mu$, I claim that the following set has measure one:
$$\{f:f(n)\text{ is eventually bounded by $n^t$ for $t>1$, but not for
$t=1$}\}$$


**Proof:**
With $\delta\_n=\sum\_{k\le n^t}\epsilon\_{nk}$ and
$\Delta\_n=\delta\_n\log 2$ and $\log=\log\_e$, we have


$$\log\prod\_{n=1}^\infty \sum\_{k\le
n^t}\frac{\log\left(1+\frac{1}{k(k+2)}\right)}{\log
2}+\epsilon\_{nk}=$$
$$\log\prod\_{n=1}^\infty \frac{ \log 2+(\log (n^t+1)-\log(n^t+2))}{\log 2}+\delta\_n =$$
$$\sum\_{n=1}^\infty \log\left[ \log 2+(\log (n^t+1)-\log(n^t+2)) +\Delta\_n \right]-\log\log 2$$


Here $\epsilon\_{nk}=\pm\frac{A}{k(k+1)}e^{-\lambda\sqrt{n-1}}$ for
certain constants $A,\lambda$.


By comparison of the negative of this series with $\sum
\frac{1}{n^t}$, where we get a constant limit, the series
converges iff $t>1$:


$$\lim\_{n\to\infty}\frac{\log\log 2-\log(\log 2+(\log (n^t+1)-\log
(n^t+2)+\Delta\_n))}{n^{-t}}=$$
$$\lim\_{n\to\infty}
\frac{(\log 2+(\log (n^t+1)-\log
(n^t+2)+\Delta\_n))^{-1}[\frac{tn^{t-1}}{n^t+1}-\frac{tn^{t-1}}{n^t+2}+\Delta'(n)]}
{-tn^{-t-1}}=$$


$$\lim\_{n\to\infty}\frac{-n^{t+1}[\frac{tn^{t-1}}{n^t+1}-\frac{tn^{t-1}}{n^t+2}+\Delta'(n)]}{t(\log 2+(\log (n^t+1)-\log
(n^t+2)+\Delta\_n))}=$$


$$\lim\_{n\to\infty}\frac{-n^{t+1}[\frac{n^{t-1}}{n^t+1}-\frac{n^{t-1}}{n^t+2}+\Delta'(n)/t]}{\log 2+(\log (n^t+1)-\log
(n^t+2)+\Delta\_n)}=$$


$$\lim\_{n\to\infty}\frac{-n^{2t}[\frac{1}{n^t+1}-\frac{1}{n^t+2}+\frac{\Delta'(n)}{tn^{t-1}}]}{\log 2+(\log (n^t+1)-\log
(n^t+2)+\Delta\_n)}=$$


$$\lim\_{n\to\infty}\frac{-n^{2t}[\frac{1}{(n^t+1)(n^t+2)}+\frac{\Delta'(n)}{tn^{t-1}}]}{\log 2+(\log (n^t+1)-\log
(n^t+2)+\Delta\_n)}=$$
$$\frac{1}{\log 2}$$


provided $n^{t+1}\Delta'(n)\to 0$.


And indeed


$$n^{t+1}\Delta'(n)=(\pm A)(\log 2)n^{t+1}\frac{d}{dn}[e^{-\lambda\sqrt{n-1}}\sum\_{k\le n^t}\frac{1}{k(k+1)}]$$


Ignoring the constant and using $f(n)=\sum\_{k\le n^t}g(k)\approx
\int\_1^{n^t}g(k)dk$, so $f'(n)\approx g(n^t)-g(1)$, we have


$$ n^{t+1} \left[ e^{-\lambda\sqrt{n-1}}[-\lambda\frac{1}{2\sqrt{n-1}}]f(n)+e^{-\lambda\sqrt{n-1}}f'(n) \right]= $$


$$ e^{-\lambda\sqrt{n-1}} n^{t+1} \left[ \left(-\lambda\frac{1}{2\sqrt{n-1}}\right)f(n)+f'(n) \right] $$


and this clearly goes to 0 as $n\to\infty$.


So $\log\prod=-\infty$ and hence $\prod=0$. This completes the proof.


