Convergence of a double sum involving prime numbers

This has been moved from math.stackexchange;

I am attempting to prove/disprove convergence of the following sum $$\lim_{n \to \infty} \frac{1}{n} \sum_{p \leq n} \sum_{k=0}^\infty \ln p \left\{\frac{n}{(p-1)p^k} \right\}$$ where $\{ x\}$ denotes the fractional part of $x$.

Let $\epsilon_n$ denote the above double sum. It's a trivial fact $\epsilon_n > 0$, since the summand contains positive terms. On the other hand, it can be shown $\epsilon_n \geq 1$. Using the facts $\{x\} + \{y\} \geq \{x+y\}$, $\{\frac{m}{n}\} \leq 1-\frac{1}{|n|}$, and $\sum_{p \leq n} \ln p \sim n$, we have

$$\epsilon_n = \frac{1}{n} \sum_{p \leq n} \ln p \sum_{k=0}^\infty \left\{\frac{n}{(p-1)p^k} \right\} \geq \frac{1}{n} \sum_{p \leq n} \ln p \left\{\frac{n}{p-1}\sum_{k=0}^\infty \frac{1}{p^k} \right\}$$

$$= \frac{1}{n} \sum_{p \leq n} \ln p \left\{\frac{np}{(p-1)^2} \right\} \geq \left(\frac{1}{n} \sum_{p \leq n} \frac{-\ln p}{(p-1)^2} +\ln p \right) \sim 1-\frac{C_n}{n}$$

where $C_n = \sum_{p \leq n} \frac{\ln p}{(p-1)^2}$. Now, the limit $C_n$ is known to be convergent (by the limit comparison test to the derivative of the prime zeta function $P(s)$, which is known to be convergent for $\mathrm{Re}(s) > 1$), so $\lim_{n \to \infty} \frac{C_n}{n} \to 0$. Thus $\epsilon_n \geq 1$. Perhaps this may not be of help, as I experience difficulty establishing an upper bound.

• Is $p\leq n$ supposed to be positive integral $p$ at most $n$, or positive integral prime values for $p$? Gerhard "Conventions Unclear For the Unconventional" Paseman, 2016.01.18 – Gerhard Paseman Jan 18 '16 at 19:38
• Oops. I missed the prime in your comment. Sorry for the confused response. I think then the sum converges, and I will see if I can help you convince others. Gerhard "Wants Confusion Replaced With Conviction" Paseman, 2016.01.18 – Gerhard Paseman Jan 18 '16 at 20:05
• For a bound on the limsup, split the sum into k=0 and k>0. For the former use {x}\leq 1 and \sum_{p\leq n} \log{p} ~ n, for the latter use {x}\leq x and \sum_p \log{p}/(p(p-1)) \leq O(1). – alpoge Jan 18 '16 at 20:51
• Yeah, that's why I split off the k=0 term and handled it separately. – alpoge Jan 18 '16 at 22:52
• And well, it looks that the limit does exist, but I am sleepy now and have to recheck this. – Fedor Petrov Jan 18 '16 at 23:05

Yes, the limit exists and equals $1-\gamma$. Summation may be taken by all $p$, it gives the same sum.

At first, we estimate the sum over $k>0$. Denote $q=(p-1)p^k$. Estimate from above $\ln p\leqslant \ln q$. We have $$\sum_{q\leqslant n} \ln q\left\{\frac nq\right\}\leqslant \ln n\sum_{q\leqslant n}1=O(\ln^2 n \cdot\sqrt{n})=o(n),$$ since if $(p-1)p^k\leqslant 1$, we have $k\leqslant \log_2 n$, $p\leqslant \sqrt{n}+1$. For $q>n$ we have $$\sum_{q>n} \ln q\cdot \frac nq=n\sum_{q\geqslant n} \frac{\ln q}{q}=o(n),$$ since the series $\ln q/q$ converges (for proving these it suffices to partition positive integers into segments $\Delta_N=[2^N,2^{N+1}-1]$, and estimate $$\sum_{q\in \Delta_N} \frac{\ln q}{q}=O\left(\frac{N}{2^N}\sum_{q<2^{N+1}} 1\right)=O(N^2\cdot 2^{-N/2})$$ as explained above.

OK, now we have to consider $k=0$, i.e., study $\sum_p \ln p \{\frac{n}{p-1}\}$. At first, it essentially (upto $o(n)$) the same as $\sum_p \ln n \{\frac{n}{p-1}\}$. Indeed, the difference is $$\sum_p \ln \frac{n}p \left\{\frac{n}{p-1}\right\}=O\left(\sum_p \ln \frac{n}p\right)=O\left(\sum_{p\leqslant n/\ln^2n} \ln \frac{n}p\right)+ O\left(\sum_{n+1\geqslant p> n/\ln^2n} \ln \frac{n}p\right)=o(n),$$ since in the first sum we have $O(n/\ln^2 n)$ summands not exceeding $\ln n$, in the second sum we have $O(\pi(n))=O(n/\log n)$ summands not exceeding $\ln\ln n$.

Well, so we have to prove that sum $S(n)=\sum_p \{\frac{n}{p-1}\}$ has asymptotics $c\cdot n/\ln n+o(n/\ln n)$ for $c=1-\gamma$. I write $\pi(n)$ instead of $n/\ln n$ as it is more natural here.

Fix $\varepsilon$ and after that large $M=M(\varepsilon)$ to be specified later. We want to determine $S(n):=\sum_p \{n/(p-1)\}$ with accuracy $\varepsilon \pi(n)$. At first, we have $$\sum_{p\leqslant n/M} \left\{\frac{n}{p-1}\right\}\leqslant \pi(n/M)<\varepsilon \pi(n)/2$$ for large enough $M$.

Now consider separately primes $p$ between $n/k$ and $n/(k+1)$ for each specific $k=1,2,\dots,M-1$. We have $$\sum_{n/k\geqslant p-1> n/(k+1)} \left\{\frac{n}{p-1}\right\}= \sum_{n/k\geqslant p-1> n/(k+1)} \frac{n}{p-1}-k\left(\pi(1+n/k)-\pi(1+n/(k+1))\right)$$ As for the first sum, we have $F(x):=\sum_{p-1\leq x} 1/(p-1)=\ln\ln x+c_0+o(1/\ln x)$, see comments on Mertens theorem here Thus $$F(n/k)-F(n/(k+1))=\ln\frac{\ln n-\ln k}{\ln n-\ln(k+1)}+o(1/\ln n)= \frac{\ln(k+1)-\ln k+o(1)}{\ln n}.$$ Next, ugly expression $k\left(\pi(1+n/k)-\pi(1+n/(k+1))\right)$ is just $\pi(n)/(k+1)+o(\pi(n))$.

Sum up by $k$, we get a limit of $\ln(M+1)-(1/2+1/3+\dots+1/M)$, this limit equals $1-\gamma$.

• +1 This is great analysis, Fedor Petrov! – Brian Diaz Jan 19 '16 at 18:39
• Hopefully, now it is not only great, but also correct. Previously I spend a lot of time to estimate the sum of $\pi(n)$ fractional parts with accuracy $o(n)$ :( :) – Fedor Petrov Jan 19 '16 at 21:36