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 $I=f(223)$? 



Sum of reciprocals of integers minus primes
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](https://mathoverflow.net/questions/212816/are-there-effective-small-intervals-in-which-primes-are-dense/212824#212824) of mine) to get
$$ \sum\_{x<p\leq x+3x^{2/3}}\log p>0.0006x^{2/3},\qquad x>\exp(8\times 10^{14}).$$


