# Fluctuating constants

Let $p_k$ be the $k$-th prime number, $\gamma$ be the Euler-Mascheroni constant and $M$ be the Meissel–Mertens and let $m$ be the integer part of $\log p_n$. We can show that

$$\sum_{r=1}^{m} \frac{e^{M + 1 + \frac{1}{2} + \ldots{} +\frac{1}{r}}}{r} = \sum_{r=1}^{n} \frac{e^{\gamma + \frac{1}{p_1} + \frac{1}{p_2} +\ldots{}+ \frac{1}{p_r}}}{p_r} + H + O\Big(\frac{1}{\ln p_n}\Big)$$

We were expecting that as $n$ increased, $H$ would approach a constant value with a small positive or negative error after each iteration. However while actually computing the constant $H$, we observed that it values fluctuates a lot between 5 and 8 as shown in the table below. We did not observe any sign of convergence.

Question:

1. Why is the constant $H$ fluctuating or rather what is its true nature?
2. Why must it fluctuate between 5 and 8?
• $1/\log(10^6)\approx 0.07$ and $1/\log(10^{20})\approx 0.03$ - so maybe the constant in the big O is just large?
– Dirk
Aug 4, 2016 at 14:09
• How is the $x$ in your table related to $p_n$? Aug 4, 2016 at 15:14
• @Wolfgang - The computations was carried all primes $\le x$. Aug 4, 2016 at 17:41

The left side of your relation is sensitive to whether $\log p_n$ is close to an integer or not, whereas the right side is not. This accounts for the fluctuations. More precisely, the contribution of an individual term on the left side is (for suitably large $r$) roughly $e^{M+\gamma}$ which is about $\exp(0.261+0.577) = 2.31$. So if $m$ changes by $1$ from a value of $p_n$ for which $\log p_n$ lies just below an integer to a value just above an integer, you'll see a jump of about $2.31$. You can already see this from your table (if I interpret the table correctly): the smallest value you show occurs around $10^{13}$ and its logarithm is $29.93$, just below an integer. The largest value you show is around $5.5 \times 10^{11}$ and the logarithm is $27.03$, just above an integer.
In any case, my opinion is that this is not an enlightening way of working with the asymptotic $$\sum_{p\le x} \frac{\log p}{p} = \log x +c + O\Big( \frac{1}{\log x}\Big).$$