Lebesgue measure of a set of irrational numbers 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? 
 A: A precise solution to this problem is known. In Khintchine's book on continued fractions it says:
$\mathbf{Theorem~30}$ Suppose that $\varphi(n)$ is an arbitrary positive function with natural argument $n$. The inequality
$$
a_n = a_n(\alpha) \geq \varphi(n)
$$
is, for almost all $\alpha$, satisfied by an infinite number of values $n$ if the series $\sum_n 1/\varphi(n)$ diverges. On the other hand, this inequality is, for almost all $\alpha$, satisfied by only a finite number of values of $n$ if the series $\sum_n 1 /\varphi(n)$ converges.
(quoting from: A. Ya. Khintchine. Continued Fractions. University of Chicago Press, 1964)
A: 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.
