Sets of unit fractions with sum $\leq 1$ 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.
 A: I think I can do slightly better. Let's look at the following sets: 
$S_1=\{A\subseteq \{\frac{e^2}{n},\dots,\frac{1}{n}\}|\sum_{x\in A}x\leq 0.9\}$.
$S_2=\{B\subseteq \{\frac{2e^2}{n},\dots,\frac{e^2}{n}\}|\sum_{x\in B}x \leq 0.1\}$.
Then obviously for any $A \in S_1, B \in S_2$, $A\cup B$ has a sum of less than one, so we have at least $|S_1|\cdot|S_2|$ sets with the desired property.
Let's try to lower bound $|S_1|$ and $|S_2|$. as for $|S_2|$, any subset of $\{\frac{2e^2}{n},\dots,\frac{e^2}{n}\}$ of size at most $\frac{0.1n}{2e^2}$ must be in $S_2$. the number of such sets is roughly $2^{H(0.1)/2e^2}=2^{0.03...}$.
Let's try to lower bound $|S_1|$. The number of subsets of $\{\frac{e^2}{n},\dots,\frac{1}{n}\}$ of size at most $0.45n(1-\frac{1}{e^2})$ is about 
$2^{H(0.45)(1-\frac{1}{e^2})}=2^{0.857...}$. For a set of such size the sum of its elements typically would be around $2\cdot0.45=0.9$. So the size of $|S_1|$ would also be $2^{0.8577}$. Multiplying $|S_1|\cdot|S_2|\geq 2^{0.887}$.
A: Let $n_0$ be the smallest number such that the sum of the reciprocals of the integers from $n_0+1$ to $n$ is $<2$.   It is easy to see that $n_0 \approx n/e^2$, since $\sum_{j>n/e^2}^{n} 1/j \approx \log n - \log (n/e^2) 
=2$.  Now for any subset $A$ of $\{n_0 +1, \ldots, n\}$ either the sum of the reciprocals of elements in $A$ or the sum of the reciprocals of its complement must be $<1$.  Therefore there are at least 
$$ 
\frac 12 2^{n-n_0} \asymp 2^{n(1-1/e^2)}
$$ 
possible sets.  My guess is that this exponent $1-1/e^2$ is correct -- note that $1-1/e^2 = 0.86466\ldots$. 
Maybe my first guess is not right!  Here's an upper bound, which gives an exponent around $0.91\ldots$ (my numerical calculations are pretty rough).  For any positive $x$, an upper bound on the quantity we want is 
$$ 
e^x \prod_{j=1}^n (1+e^{-x/j}). 
$$ 
To see this, just expand out the product and terms with sum of reciprocals less than $1$ will contribute at least $1$, and the rest are positive. Now choose $x$ so as to minimize the above (a standard idea, known in analytic number theory as Rankin's trick).  
Calculus shows that one must choose $x$ so that 
$$
1= \sum_{j=1}^{n} \frac 1j \frac{1}{1+e^{x/j}}. 
$$
It is natural to guess that $x$ is of the shape $\alpha n$ for a 
constant $\alpha$, and then for large $n$ the condition on $\alpha$ becomes 
$$ 
1= \int_0^1 \frac{1}{1+e^{\alpha/y}} \frac{dy}{y} = \int_1^\infty \frac{1}{1+e^{\alpha y}} \frac{dy}{y}. 
$$ 
If I calculated right, this gives $\alpha \approx 0.1273$.  For this choice of $\alpha$ (and so $x$), one obtains the bound (approximately) 
$$ 
\exp\Big(n\Big( \alpha + \int_0^1 \log (1+e^{-\alpha/y}) dy\Big)\Big), 
$$
which seems to be about 
$$ 
\exp(-.631n) \approx 2^{0.911n}.
$$ 
(I won't swear to the numerics -- someone should check.)  
My second guess is that the upper bound is tight (and I think this could be proved with some effort).  The idea is to choose $j$ to be in your set with probability $1/(1+\exp(x/j))$ with the same $x$ as in the upper bound.  The expected value of $1/j$ with this distribution is $1$, by the choice of $x$. An entropy calculation for this distribution then gives the exponent.  (More generally, in all the situations I know, the Rankin upper bound is pretty close to optimal.)
A: 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
A: Let $R > 1$ and $\lambda \in \mathbb{R}$ be such that
$$
\int_{1}^R \mathrm{tanh}(\frac{\lambda x}{2}) \frac{d x}{x} =  \log R -2.
$$
Then standard techniques in large deviation theory yield
$$
\frac{1}{n} \log | \{ I \subseteq [n R^{-1},n] | \sum_{i \in I} \frac{1}{i} \leq 1 \} | \longrightarrow \int_{1}^R \phi(\lambda x)\frac{d x}{x^2} 
$$
where $\phi(u) = \log2 +\log \cosh (\frac{u}{2}) - \frac{u}{2} = \log(1 + e^{-u})$. For $R = e^2$ one has $\lambda =0$ and the limit is $(1 - e^{-2})\log 2$ : this recovers Lucia's first result.
Moreover, letting $R$ tend to infinity, then $\lambda$ tends to Lucia's constant $\alpha$, and the limit integral matches Lucia's upper bound. In particular Lucia's "second guess" was correct.
EDIT: Some details about the said "standard techniques". Let $(X_i)_i$ be independent Bernoulli random variable ($= \pm 1$ with probability $1/2$), and let $(t_i)_{i=1}^n$ be positive real numbers lying in a fixed interval $(-R,R)$. We are going to study the probability that $S = \sum_{i=1}^n t_i X_i$ is $\geq c$, for some $c \in \mathbb{R}$. Let $\lambda \in \mathbb{R}$ be such that
$$
\frac{1}{n} \sum_{i=1}^n t_i \tanh(\lambda t_i) = c.
$$
This amounts to say that $E[\frac{1}{n} S e^{\lambda S}] = c E[e^{\lambda S}]$. Such a $\lambda$ exists as soon as $|c| < \frac{1}{n} \sum_{i=1}^n t_i$. One checks that
$$
E[(\frac{1}{n} S)^2 e^{\lambda S}]E[e^{\lambda S}]^{-1} = c^2 + \frac{1}{n^2} \sum_{i=1}^n t_i^2(1- \tanh(\lambda t_i)^2)
$$
so that the variance of $\frac{1}{n} S$ w.r.t. the measure weighted by $e^{\lambda S} E[e^{\lambda S}]^{-1}$, is at most $R^2 n^{-1}$. In particular
$$
E[ \mathbb{1}_{\frac{1}{n} S \in [c,c+\epsilon]} e^{\lambda S}]  E[e^{\lambda S}]^{-1}  = \frac{1}{2} + O(R^2 \epsilon^{-2} n^{-1}).
$$
Thus
$$
P( S \geq nc) \geq e^{-n \lambda (c+\epsilon)}E[ \mathbb{1}_{\frac{1}{n} S \in [c,c+\epsilon]} e^{\lambda S}] = e^{-n \lambda (c+\epsilon)} E[e^{\lambda S}] \left( \frac{1}{2} + O(R^2 \epsilon^{-2} n^{-1}) \right)
$$
On the other hand
$$
P( S \geq nc) \leq e^{-n \lambda c} E[e^{\lambda S}]
$$
so that
$$
\frac{1}{n} \log P( S \geq nc) = - \lambda c + \frac{1}{n} \log E[e^{\lambda S}] + o_{R}(1)
$$
For the application above, just use $t_i = \frac{n}{2i}$ (indexed by $i$ between $n R^{-1}$ and $n$) and $c = \frac{1}{2} \sum_{n R^{-1} < j \leq n} \frac{1}{j} - 1$.
