# Better error bounds for partial sums of reciprocals of primes?

One of Mertens' theorems gives that

$$\sum_{ p \text{ prime,} p \leq k } 1/p - \log{\log{k}} = B + E(k)$$

where $B$ is a constant near $0.26$ in value and $E(k)$ is an error term whose size is dominated by something close to $4/\log{k}$, when $k$ is large enough to make the sum meaningful.

I want to work with partial sums of the above with $j \lt p \leq k$, so that I can say the partial sum of the reciprocals of primes greater than $j$ and at most $k$ is $\log{(\log{k}/\log{j})} + E(j,k)$ where $j$ is not too small (perhaps $j \gt 3$ or $5$), $k$ not too large, say $j^\alpha \lt k \lt j^\beta$ where often $1 \lt \alpha \lt \beta \lt e$, and $E(j,k)$ is comfortably small. Unfortunately $4[1/\log{k} + 1/\log{j}]$ looks too big for me; I am hoping to have (for large enough $j$) $E(j,k)$ bounded by something that is $O(1/j)$ or better.

I have access to Mark Villarino's treatment of Mertens' theorem. As of 2005, it seems $E(k)$ is no better than $O(1/\log{k}^2)$ I also hope to obtain recent work of Pintz and Diamond on oscillations ia related formula which is Mertens product formula, but I do not see yet how it will me help me with this formula.

As I am still a tyro at number theory, I don't even know how realistic my hopes are for $E(j,k)$ to be $O(1/j)$. Can someone who is familiar with the recent literature give me a guide post? Either references or heuristics showing what sort of bounds to expect for $E(j,k)$ or even $E(k)$ would be welcome.

UPDATE 2012.07.24 I want to acknowledge the contributions of joro, Christian Elsholtz, and Eric Naslund. joro and Eric helped me realize that expecting $O(x^{-1/2})$ error even conditionally is expecting a bit much, and Christian helped me realize that Dusart still has some nice unconditional refinements. I will likely accept Christian's answer, but not before I do some computations of my own. In particular, Rosser and Schoenfeld have in Theorem 20 of their 1962 paper on functions relating to primes a nice difference of $2/(x^{1/2}\log x)$ which is valid for $1 \lt x \lt 10^8$ between lower and upper estimates for the sum of interest, and I may end up using or refining that estimate in combination with Dusart's results for larger $x$ for my own nefarious purposes. I am hoping in particular that the oscillations will be slow enough that my desired partial sums from $j$ to $j^\alpha$ will have very small error. END UPDATE 2012.07.24

Gerhard "Ask Me About System Design" Paseman, 2012.07.19

• This looks interesting. Have you obtained the estimate you wanted for the partial sums from $j$ to $j^{\alpha}$? – kodlu Feb 25 '16 at 22:30
• @kodlu I got something close enough for my purposes. arxiv.org/abs/1311.5944 points to a writeup which has a Lemma on page 14 where I used this partial sum. I was able to make do with a small but bounded away from zero error which I used to prove an assertion that held for numbers with "enough" prime factors. I did not need something as small 1/(x^{1/2}) error, and the end result was instructive for me but far from groundbreaking. Gerhard "Perhaps Also Far From Noteworthy" Paseman, 2016.02.25 – Gerhard Paseman Feb 26 '16 at 0:44

## 3 Answers

In case you also want computationally explicit bounds look at Theorem 6.10 of:

Pierre Dusart: Estimates of Some Functions Over Primes without R.H. http://arxiv.org/abs/1002.0442

Probably the proof is for you more interesting than the Theorem itself. On the one hand side it shows (as Eric did) the connection with the error term of the prime number theorem. On the other hand, with the constants detailed in Theorem 5.2 you can construct explicit bounds of the type. If $x>x_k$, then $\vert \sum_{p < x} \frac{1}{p}- B - \log \log x \vert < \frac{\eta_k /k}{(\log x)^k} + \frac{\eta_k (1 + \frac{1}{k+1})}{ (\log x)^{k+1}}$.

The asymptotic bounds given by the error term of the prime number theorem are of course stronger. (Explicit bounds of the type that Eric mentioned exist, but are probably not useful for computations).

• Explicit, unconditional, and computationally useful bounds are what I desire. This answer comes closest and is long overdue for acceptance. Gerhard "Thanks You For This Answer" Paseman, 2018.02.06. – Gerhard Paseman Feb 6 '18 at 17:10

Under Riemann hypothesis you can get better bound.

Sharper bounds for the Chebyshev functions $\theta (x)$ and $\psi (x)$. II Lowell Schoenfeld

http://www.ams.org/journals/mcom/1976-30-134/S0025-5718-1976-0457374-X/

(6.21) p.4 $$\left|\sum_{p \le x}\frac1p-\log\log{x}-B\right|< \frac{3 \log{x} +4}{8 \pi \sqrt{x}}$$

if $13.5 \le x$.

Up to a factor of logarithm, $E(x)$'s oscillation has an amplitude which is of the same magnitude as that of $\frac{1}{x}\left(\psi(x)-x\right)$, that is the error in the prime number theorem. Specifically $$E(x)=O\left( e^{-c(\log x)^{3/5-\epsilon}}\right)$$ unconditionally, and $$E(x)=O\left( x^{-\frac{1}{2}+\epsilon}\right).$$ under RH.

Proof: Let $W(x)=\pi(x)-\text{li}(x)$ be the error term in the prime number theorem. Then

$$\sum_{p\leq x}\frac{1}{p}=\int_{2}^{x}\frac{1}{t}d\left(\pi\left(t\right)\right)=\int_{2}^{x}\frac{1}{t\log t}dt+\int_{2}^{x}\frac{1}{t}dW(t).$$

Integration by parts yields

$$\int_{2}^{x}\frac{1}{t}dW(t)=\frac{W(t)}{t}\biggr|_{2}^{x}+\int_{2}^{x}\frac{W(t)}{t^{2}}dt.$$

Since

$$\int_{2}^{x}\frac{W(t)}{t^{2}}dt=\int_{2}^{\infty}\frac{\pi(t)-\text{li}(t)}{t^{2}}dt+O_{\epsilon}\left(e^{-c\left(\log x\right)^{\frac{3}{5}-\epsilon}}\right),$$

it then follows that $$\sum_{p\leq x}\frac{1}{p}=\log\log x+b+O_{\epsilon}\left(e^{-c\left(\log x\right)^{\frac{3}{5}-\epsilon}}\right),$$

where $b=\int_{2}^{\infty}\frac{\pi(t)-\text{li}(t)}{t^{2}}dt+\frac{W(2)}{2}-\log\log2.$

• I think the exponent $2/3$ should be $3/5$ (Vinogradov and Korobov. Some "caveat" is necessary, but the $\varepsilon$ cares for it). – Christian Elsholtz Jul 20 '12 at 8:57
• @Christian: Thank you for the correction! I am not sure why I was thinking $2/3$. (I put the ϵ rather than the more precise $(\log \log x)^\frac{-1}{5}$ term because I didn't want the exponential to be more cramped then it already was - but I am not sure if that is good reasoning...) – Eric Naslund Jul 20 '12 at 10:54
• @EricNaslund:In the comment above, do you mean the error term can be written as big oh of $$\exp \left\{-c\left(\frac{(\log x)^3}{\log\log x}\right)^{1/5}\right\}$$ ? – kodlu Feb 25 '16 at 22:45