TITLE: Sets of unit fractions with sum $\leq 1$
QUESTION [41 upvotes]: Consider a set of fractions $\left\{1, \frac{1}{2}, \frac{1}{3}, \ldots, \frac{1}{n}\right\}$. How many subsets of this set have sum at most 1? I'm interested in the asymptotics of this number.
Clearly, any subset of $\left\{\frac{1}{\lceil n/2 \rceil}, \ldots, \frac{1}{n}\right\}$ works, hence the answer is $\Omega(2^{n / 2})$. Can we show $\Omega(2^{\beta n})$ for $\beta > \frac{1}{2}$? Can we determine $\beta$ exactly? From numeric estimates of OEIS sequence $\beta$ seems to at least $0.88$ (link to the sequence and correction of the estimate due to Max Alekseyev).
The question arose while I was thinking about upper bounds for this question. Clearly, every divisibility antichain $I$ of $[n]$ must satisfy $\sum_{x \in I} \lfloor\frac{n}{x}\rfloor \leq n$, which is a very similar condition.
Post-mortem: while I accepted Lucia's answer (simply because it was the first to contain the correct answer and some reasoning to why it is correct), the whole discussion here is very valuable. Be sure to also check out js21's answer with an approach based on large deviations method, and RaphaelB4's answer for a more off-the-ground explanation of the method. In a comment Jay Pantone shared a link to a paper on series analysis, in particular, the differential approximation method allows to obtain the same answer with high precision and is, without doubt, a great practical tool. Kudos to all of you guys! What a great day to learn.

REPLY [18 votes]: The bound proposed by Lucia is correct. I add some detail and I stress that It is an application of "large deviation" theorem which is very very standard.
Set the following independent Bernoulli random variable defined as $$X_i=\begin{cases} 0 \text{ with } p=1/2\\ \frac{1}{i} \text{ with } p=1/2\end{cases}$$
Then we have the number of subset is given by
$$ 2^n \mathbb{P}(\sum_{i=1}^n X_i \leq 1)$$
And we can then adapte the proof of the well known Cramer theorem. 
For a lower bound 
$$\mathbb{P}(\sum_{i=1}^n X_i\leq 1)=\mathbb{P}(e^{-x\sum_{i=1}^n X_i}\geq e^{-x})\leq \frac{\mathbb{E}(e^{-x\sum_{i=1}^n X_i})}{e^{-x}}$$
which give because of the independence of $X_i$ the formula that Lucia have already stated (I didn't know it is called Rankin bound)
$$\mathbb{P}(\sum_{i=1}^n {X}_i \leq 1)\leq e^{x} \prod_i^n (e^{-\frac{x}{i}}+1)/2^n$$ 
There exists $x_0$ which minimise the right part.
Then introduce
$$\tilde{X}_i=\begin{cases} 0 \text{ with } p=\frac{1}{1+e^\frac{-x_0}{i}}\\ \frac{1}{i} \text{ with } p=\frac{e^\frac{-x_0}{i}}{1+e^\frac{-x_0}{i}} \end{cases}$$
Remark that 
$$ \mathbb{E}(\sum_i \tilde{X}_i)=\frac{\mathbb{E}(\sum_i X_i e^{-x_0\sum_{i=1}^n X_i})}{\mathbb{E}(e^{-x_0\sum_{i=1}^n X_i})}=\partial_x [ln(\mathbb{E}(e^{-x_0\sum_{i=1}^n X_i}))]_{x=x_0}$$
But because $x_0$ is a minimum, $\partial_x [ln(\mathbb{E}(e^{-x_0\sum_{i=1}^n X_i}))-ln(e^{-x})]_{x=x_0}=0$ and therefore $\mathbb{E}(\sum_i \tilde{X}_i)=1$. But because of convexity for any $\epsilon>0$ changing $\tilde{X}$ by $\tilde{X}^\epsilon$ by replacing $x_0$ by $x=x_0+\epsilon n$ in the definition give $\mathbb{E}(\sum_i \tilde{X}_i)\leq 1-\delta(\epsilon)$ with $\delta(\epsilon)>0$. We have then that $\mathbb{P}(\sum\tilde{X}^\epsilon_i\leq 1)\geq \delta(\epsilon)$
We can then state the lower bound 
$$ \delta(\epsilon)\leq \mathbb{E}(1_{\sum(\tilde{X}_i^\epsilon)\leq 1})\leq \frac{\mathbb{E}(1_{\sum X_i \leq 1} e^{-x \sum_i X_i})}{\mathbb{E}(e^{-x \sum_i X_i})} \leq \frac{\mathbb{E}(1_{\sum X_i \leq 1} )e^{-x}}{\mathbb{E}(e^{-x \sum_i X_i})}$$ And to conclude
$$ \delta(\epsilon)\frac{\mathbb{E}(e^{-x \sum_i X_i})}{e^{-x}}\leq \mathbb{P}(\sum X_i \leq 1)$$
and therefore
$$\lim \frac{1}{n}\log(\frac{\mathbb{E}(e^{-x \sum_i X_i})}{e^{-x}})\leq \lim \frac{1}{n}\log(\mathbb{P}(\sum X_i \leq 1))$$
which is true for any $\epsilon>0$.
This is the end of the proof that the bound is tight