Can every positive real be written as the sum of a subsequence of dot dot dot I answered this thing Infinite sum of prime reciprocals and now wonder what happens if we do not have such a strong condition as Bertrand's postulate. i have been fiddling with this, not sure either way.
Given a sequence $a_1 > a_2 > a_3 \cdots$ of strictly decreasing positive reals such that
$$ a_i \rightarrow 0 \; \; \; \mbox{but} \; \; \sum a_i = \infty,  $$
can every positive real number be expressed as the sum of a subsequence of the $a_i?$ The main thing is that we are not given any upper bound on $a_n / a_{n+1}.$ For the reciprocals of the primes, we had an upper bound of $2.$
Note that this is subtler than the thing about rearranging a strictly alternating conditionally convergent series to get anything you specify. That is a matter of overshooting with positive terms, then undershooting with negative terms, back and forth. This one is a little different. 
I think what I want is a careful proof of this: given two positive real numbers $B<C,$ we can find a finite subsequence of the $a_n$ with sum between $B$ and $C.$

The key is that the conditions guarantee that, given $\varepsilon>0$, we can find a subsequence $\{b_n\}$ with $0<b_n<\varepsilon$ and $\sum b_n=\infty$. 
So, given $r>0$, choose $a_{n_1},\ldots,a_{n_{k_1}}$ with $0<a_{n_j}<1/2$ and $$r-1/2\leq a_{n_1}+\cdots+a_{n_{k_1}}<r.$$
Next we choose $a_{n_k+1},\ldots,a_{n_{k_2}}$ with $0<a_{n_j}<1/3$ and
$$r-1/3\leq a_{n_1}+\cdots+a_{n_{k_2}}<r$$
(maybe none are needed). Repeating this, we eventually will have $a_{n_1},\ldots,a_{n_{k_m}}$ such that 
$$
r-1/m<a_{n_1}+\cdots a_{n_{k_m}}<r.
$$
