Roughly how many subsets of $[n]:=\{1,2,\ldots,n\}$ have the property that the sum of the reciprocals of the members in that subset is less than $x$? Let $\ n\ $ be a large positive integer, and let $\ 0 < x < \displaystyle\sum_{k=1}^{n} \frac{1}{k}.\ $ Approximately how many subsets of $\ [n]:= \{ 1,2,\ldots,n\}\ $ have the property that the sum of the reciprocals (of the members in that subset) is less than $x$ ?
 A: This is just a partial answer, which is too long for a comment.
The question can be equivalently formulated as: given iid Bernoulli random variables $\varepsilon_k$, $k\ge1$, with $\mathrm P(\varepsilon_k = 1) = \mathrm P(\varepsilon_k = 0) = \frac12$, what is the probability that
$$
S_n := \sum_{k=1}^n \frac{\varepsilon_k}k <x?
$$
Writing
$$
S_n \approx \frac{\log n}2 + \sum_{k=1}^n \frac{\varepsilon_k-1/2}{k}
$$
and observing that the series $\zeta :=\sum_{k=1}^\infty \frac{\varepsilon_k-1/2}{k}$ converges by Kolmogorov's one series theorem, we get that $\mathrm{P}(S_n<x)\approx 1$ for $x\gg \frac{\log n}2$, and $\mathrm{P}\left(S_n<\frac{\log n}2 +t \right)\to  \mathrm P(\zeta<t)$ for every $t\in \mathbb R$. The distribution of $\zeta$ is studied in Random Harmonic Series by Byron Schmuland.
So the most interesting case is $x\ll \frac{\log n}2$, where we do know that $\mathrm{P}(S_n<x)\approx 0$, but this does not tell a lot about the number of sums (which can be anything negligible compared to $2^n$).
A: This is mainly a comment which is too long to fit as such.
Since the harmonic numbers themselves satisfy that
$$ H_n = \log(n)+ \gamma + o(n) $$
I agree with zhoraster that the interesting case must be $x<< \log(n)$.
To try and understand zhorasters $\zeta$-variable I used the following Python script
def brute_count(n):
        results=[]
        for sample in subset_generator(n):
            results.append(sum([1/x for x in sample]))
        results.sort()
        return results

def subset_generator(n):
    universe = list(range(1,n+1))
    subsets=[[]]
    for elt in universe:
        subsets += [s+[elt] for s in subsets]
    return subsets

I get a sorted list of all conceivable sums arising as described in the question for reasonably sized $n$. Now if the $i$-th element in this list has size $y$ that would mean that there are $i$ partial sums with value less than or equal to $y$. In other words the plot
ln14=brute_count(14)  
plt.plot(ln14,range(1,16385));plt.savefig("image.png") 

corresponds to the function
$$ x \ \mapsto \ \#\{ I \in \{1,\ldots,n\} \ | \ \sum_{l \in I} \frac{1}{l} < x \ \}.$$
for $n=14$. For $n$ going to $\infty$ this function should - by the law of large numbers - converge pointwise to
$$ x \ \mapsto \ 2^n\cdot P(S_n<x)$$
The plot is as follows 
If we normalize our function as follows
>>> plt.plot([(l-h14/2)/(h14/2) for l in ln14],[l*1/16384 for l in range(1,16385)])

(where h14 is $H_{14}$) we get

Up to a few additional normalizations this should correspond to zhorasters $\zeta$-variable.
I have tried to compare this to a few standard cumulative distribution functions appropriately normalized, but didnt really convince myself. Eitherway I think it should be possible to calculate the moments of the distribution corresponding to l14 numerically, and if these can be recognized as being very close to rational numbers of small height or something similar, then perhaps you can make a conjecture about the distribution $S_n$ in the interesting range, redo the calculation for larger $n$ and then test it.
