Does $\lim_{x \to \infty} \ln(\frac{x}{m}) \sum_{p \le x} \frac{1}{\ln p}$ converge? Does $$\lim_{x \to \infty} \ln\left(\frac{x}{m}\right)  \sum_{p \le x} \frac{1}{\ln p}$$ converge? Here, $p$ ranges over the primes, and $m$ is the largest prime less than $x$. I think that $\frac{x}{m}$ tends to infinity, but the sum tends to $0$; I'm not sure about the sum at all.
 A: Let $p_1, p_2, \cdots$ the sequence of all primes in increasing order, let $g_n = p_{n+1}-p_n$ the sequence of prime gaps. Let $\pi(t)=\sum_{p\le t} 1$ be the prime counting function. 
By partial summation and the Prime Number Theorem, we have
$$
\begin{align}
\sum_{p\leq x} \frac1{\ln p} &= \int_{2-}^x \frac1{\ln t} d\pi(t) \\
&=\frac{\pi(t)}{\ln t} \Bigg\vert_{2-}^x + \int_{2-}^x \frac{\pi(t)}{t\ln^2 t}dt\\
&=\frac{\pi(x)}{\ln x}+O\left(\int_2^x \frac{dt}{\ln^3 t}\right)\\
&=\frac x{\ln^2 x}+O\left(\frac x{\ln^3 x}\right).
\end{align}
$$
If we take $x=p_n+1$, then 
$$
\ln\left( \frac xm\right)=\ln\left( \frac{p_n+1}{p_n}\right)=\frac1{p_n}+O\left(\frac 1{p_n^2}\right).
$$
Thus, we have 
$$
\begin{align}
\ln\left(\frac xm \right) \sum_{p\le x}\frac 1{\ln p} &=O\left(\frac 1{\ln^2 p_n}\right) \rightarrow 0 \textrm{ as }n\rightarrow\infty.
\end{align}
$$
This shows that 
$$
\liminf_{x\rightarrow\infty}\left[ \ln\left(\frac xm \right) \sum_{p\le x}\frac 1{\ln p}\right]=0.
$$
On the other hand, we take $x=p_{n+1}$. Then, we have
$$
\ln\frac xm = \frac {p_n+g_n}{p_n} = \frac{g_n}{p_n} + O\left( \frac{g_n^2}{p_n^2} \right).
$$Thus,
$$\begin{align}
\ln\left(\frac xm \right) \sum_{p\le x}\frac 1{\ln p}&=\frac{g_n p_{n+1} }{p_n\ln^2 p_{n+1}}+O\left(\frac {g_n^2p_{n+1}}{p_n^2 \ln^2 p_{n+1} } \right)+O\left(\frac{g_np_{n+1} }{p_n\ln^3 {p_{n+1}}}\right)\\
&=\frac{g_n p_{n+1} }{p_n\ln^2 p_{n+1}}\left(1+O\left(\frac{g_n}{p_n}\right)+O\left(\frac1{\ln p_{n+1}}\right)\right)
\end{align}.$$
Since $1+O\left(\frac{g_n}{p_n}\right)+O\left(\frac1{\ln p_{n+1}}\right)$ converges to $1$ by the Prime Number Theorem, the main term $\frac{g_n p_{n+1} }{p_n\ln^2 p_{n+1}}$ will be of our interest. Thus, it is enough to consider $\frac{g_n}{\ln^2 p_n}$. 
The Riemann Hypothesis implies that 
$$
g_n=O(\sqrt{p_n}\ln p_n). 
$$
This is not enough for the convergence of $\frac{g_n}{\ln^2 p_n}$. 
Cramer's conjecture states that 
$$
\frac{g_n}{\ln^2 p_n}=O(1).
$$
However, this is still not enough for convergence. Therefore, we know that the $\liminf$ is zero, but we do not know if $\limsup$ exists. 
A: We have that
1) $\sum \limits_{p \leq x} \frac{1}{p} \leq \ln \ln x + B +\frac{1}{\ln^2 x}$ for large number $x$,where $B \approx 0.261497$ is the Meissel–Mertens constant.
2) There is a prime between $x(1-\frac{1}{\ln^2 x})$ and $x$,(the proof by Dusart is simple since $\theta(x)$ "jumps" change value when it hits a prime so if $\theta(x) - \theta(x(1-\frac{1}{\ln^2 x})) >0$ then we have a prime between them and we are done :), for large number $x$.
Now by the above try to maximize :
$\lim \limits_{x \to \infty} \ln \frac{x}{m}  \sum \limits_{p \leq x} \frac{1}{p} = \lim \limits_{x \to \infty} (\ln x-\ln m)(\ln \ln x+ B+\frac{1}{\ln^2 x})$
we maximize that by given $m$ lower value so $m=x(1-\frac{1}{\ln^2 x})$
So we arrive at $\lim \limits_{x \to \infty} (\ln x-\ln x-\ln(1-\frac{1}{\ln^2x}))(\ln \ln x+ B+\frac{1}{\ln^2 x})$
because $-\log \left(1-\frac{1}{t}\right)\leq \frac{1}{t-1}$ for $t>1$.
we have that $\lim \limits_{x \to \infty} \frac{1}{-1+\ln^2 x}(\ln \ln x+ B+\frac{1}{\ln^2 x})$
From here its easy to see that $\lim \limits_{x \to \infty} \ln \frac{x}{m}  \sum \limits_{p \leq x} \frac{1}{p} =0$
