Let $\alpha>0$ be any positive real number and let $k\ $ be any positive integer satisfying $k > \frac{1}{\alpha}\ $
Consider the integer sequence define by: $a(1)=k;\ $ and for $n\geq 1,\ a(n+1)\ $ is the least integer $>a(n)$ such that $\displaystyle\sum_{i=1}^{n+1} \frac{1}{a(i)} < \alpha.$
What is the (asymptotic) growth rate of $a(n)$?
I would imagine exponential or maybe even along the lines of factorial, but I do not know how to study such sequences, although obviously I am interested in them.
It's possible that this has to do with continued fractions, although I have never really understood continued fractions all that well, so I'm not really sure.
Thanks in advance.
Let $(a(n))_{n\in\mathbb N}$ be a sequence of natural numbers. Given a number $\alpha>0$, we shall call the sequence $(a(n))_{n\in\mathbb N}$ $\alpha$-admissible, if it is build given $\alpha$ and $a(1)$, as it is described in the question. We shall call the sequence $(a(n))_{n\in\mathbb N}$ admissible, if it is $\alpha$-admissible for some $\alpha>0$. We can characterize all $\alpha$-admissible sequences as follows.
Proposition. For any real $\alpha>0$ and any increasing sequence $(a(n))_{n\in\mathbb N}$ of natural numbers the following conditions are equivalent:
1)) The sequence $(a(n))_{n\in\mathbb N}$ is $\alpha$-admissible.
2)) $\sum_{i=1}^\infty \frac 1{a(i)}=\alpha$ and for each natural $n$, either $a(n+1)=a(n)+1$ or $\frac{1}{a(n+1)-1}\ge \sum_{i=n+1}^\infty \frac 1{a(i)}$.
3)) $\sum_{i=1}^\infty \frac 1{a(i)}=\alpha$ and there exists a natural number $N$ such that for each natural $n$ we have $a(n+1)=a(n)+1$ if $n\le N$, and $a(n+1)>a(n)+1$ and $\frac{1}{a(n+1)-1}\ge \sum_{i=n+1}^\infty \frac 1{a(i)}$, if $n>N$.
Proof.
$(1)\Rightarrow (2)$ We have $\sum_{i=1}^\infty \frac 1{a(i)}=\alpha$, because otherwise, since the harmonic series $\sum_{i=1}^\infty \frac 1i$ diverges, at some step $i$ we had to choose a smaller value of $a(i)$. Next, for each natural $n$, the choice of $a(n+1)$ ensures that either $a(n+1)=a(n)+1$ or $\frac{1}{a(n+1)-1}\ge \sum_{i=n+1}^\infty \frac 1{a(i)}$.
$(2)\Rightarrow (3)$ Since the series $\sum_{i=1}^\infty \frac 1{a(i)}$ and the harmonic series diverges, there exists $$N=\max\{N': a(n+1)=a(n)+1\mbox{ for each natural }n\le N'\}.$$ Suppose for a contradiction that the set $M=\{n\in\mathbb N: n>N \mbox{ and } a(n+1)=a(n)+1\}$ is nonempty. Let $m=\min M$. Then $$\frac{1}{a(m)-1}>\frac{1}{a(m)}+\frac{1}{a(m+1)}=\frac{1}{a(m)}+\frac{1}{a(m)+1}.$$ That is $(a(m)-1)^2<2$, so $a(m)=2$, a contradiction.
$(3)\Rightarrow (1)$ Easy to check. $\square$
It follows that the growth rate of an admissible sequence $(a(n))_{n\in\mathbb N}$ can be arbitrary big. Namely, it suffices to require for each $n\in\mathbb N$ that $a(n+1)\ge a(n)+2$ and $$\frac{1}{a(n+1)(a(n+1)-1)}\ge \sum_{i=n+2}^\infty \frac 1{a(i)}.$$
On the other hand, any admissible sequence $(a(n))_{n\in\mathbb N}$ satisfies the latter condition for all sufficiently big $n$. Then for those $n$ we have $a(n+2)\ge a(n+1)(a(n+1)-1)+1,$ so $a(n+2)-\frac 12> \left(a(n+1)-\frac 12\right)^2$. Therefore we have $a(n+k+1)-\frac 12> \left(a(n+1)-\frac 12\right)^{2^k}$ for each natural $k$.
This lower bound of the growth rate of an admissible sequence is rather tight. Indeed, for any number $N$ we can construct a sequence $(a(n))_{n\in\mathbb N}$ putting $a(n)=n$ for any natural $n\le N$ and $a(N+k+1)=(N+2)^{2^k}$ for each nonnegative integer $k$. It is easy to check that the sequence $(a(n))_{n\in\mathbb N}$ is admissible.