# Order of magnitude of $\sum \frac{1}{\log{p}}$

Question: What is the order of magnitude of the following sum?

$$\sum_{\substack{p<n\\\text{p prime}}} \frac{1}{\log{p}}$$

$$\sum_{\substack{p<n\\\text{p prime}}} \frac{1}{\log{n}} \leq \sum_{\substack{p<n\\\text{p prime}}} \frac{1}{\log{p}} \leq \sum_{p<n} \frac{1}{\log{p}},$$

we have that, for some constants $c_1,c_2$,
$$c_1\frac{n}{\log^2{n}} \leq \sum_{\substack{p<n\\\text{p prime}}} \frac{1}{\log{p}} \leq c_2 \frac{n}{\log{n}}.$$

Here, the asymptotics on the left hand side came from the prime number theorem, and on the right hand side from the asymptotic expansion of the logarithmic integral function.

• The upper bound can be improved to $O(n/\log^2n)$ for free, simply by splitting the summation range into the two subintervals $p<\sqrt n$ and $\sqrt n<p<n$. (For the second subinterval, each summand is $O(1/\log n)$, and the number of summands is $O(n/\log n)$ by the prime number theorem.
– Seva
Nov 15 '13 at 20:11

The contribution of the primes $p\leq n/\log ^3 n$ is clearly $O(n/\log^3 n)$. For the remaining primes we have $$\log n-3\log\log n <\log p<\log n,$$ $$\frac{1}{\log p}=\frac{1}{\log n}+O\left(\frac{\log\log n}{\log^2 n}\right),$$ so by the Prime Number Theorem their contribution is $$\left(\frac{n}{\log n}+O\left(\frac{n}{\log^2 n}\right)\right)\left(\frac{1}{\log n}+O\left(\frac{\log\log n}{\log^2 n}\right)\right)=\frac{n}{\log^2 n}+O\left(\frac{n\log\log n}{\log^3 n}\right).$$ Altogether we see that $$\sum_{p<n} \frac{1}{\log{p}} =\frac{n}{\log^2 n}+O\left(\frac{n\log\log n}{\log^3 n}\right).$$
Better bounds can be obtained by partial summation, namely $$\sum_{p\leq n} \frac{1}{\log{p}} = \frac{\pi(n)}{\log n}+\int_2^n\frac{\pi(x)}{x\log^2 x}dx.$$