Sum of reciprocals of integers minus primes For any integer $m>2$, let $P_m$ be the set of primes less than $m$, and let
$$
f(m) = \sum\limits_{p \in P_m} \frac{1}{m-p}.
$$
For example, $f(3)=\frac{1}{3-2}=1$, $f(4)=\frac{1}{4-2}+\frac{1}{4-3}=\frac{3}{2}$, and so on.
The question is to estimate $I=\inf\limits_{m>2} f(m)$.
A simple Mathematica calculation shows that $f(m)\geq f(223)\approx 0.60178$ for all $m$ up to $10,000$. It is true that $I>0$? Is $I>0.5$? Is $I=f(223)$? 
 A: We have $f(m)>0.53899$ for $m$ sufficiently large. Under the Riemann hypothesis, we even have $f(m)>0.69314$ for $m$ sufficiently large, which would also imply that $f(m)$ attains a minimum.
If $m-1$ is a prime, then clearly $f(m)\geq 1$. Otherwise we have
$$f(m)=\int_0^{m-1}\frac{d(\pi(x)-\pi(m))}{m-x}=\frac{\pi(m)}{m}+\int_0^{m-1}\frac{\pi(m)-\pi(x)}{(m-x)^2}\,dx.$$
Fix a constant $7/12<c<1$. By a result of Huxley (1972), we have 
$$\pi(m)-\pi(x)\geq(1+o(1))\frac{m-x}{\log(m-x)}\qquad\text{for}\qquad x<m-m^c.$$
Here, $o(1)$ is meant as $m$ tends to infinity. Therefore,
$$f(m)\geq(1+o(1))\int_0^{m-m^c}\frac{1}{(m-x)\log(m-x)}\,dx=\log(1/c)+o(1).$$
As $\log(12/7)>0.53899$, my first claim is proved. Under the Riemann hypothesis, we can take any $1/2<c<1$, and then $\log 2>0.69314$ justifies my second claim.
Added. By a variant of the above argument, one can get an explicit lower bound for all $m$ (rather than for $m>m_0$) using an explicit prime number theorem in short intervals (in place of Huxley's result). For example, one can use the work of Dudek (as explained in this earlier MO post of mine) to get
$$ \sum_{x<p\leq x+3x^{2/3}}\log p>0.0006x^{2/3},\qquad x>\exp(8\times 10^{14}).$$
