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

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

Additional information: Since

$$ \sum_{p<n}_{p\ \ prime} \frac{1}{\log{n}} \leq \sum_{p<n}_{p\ \ prime} \frac{1}{\log{p}} \leq \sum_{p<n} \frac{1}{\log{p}}. $$

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

Where 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.

(See: http://en.wikipedia.org/wiki/Logarithmic_integral_function#Asymptotic_expansion)

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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
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1 Answer

up vote 14 down vote accepted

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 $$ \sum_{p\leq n} \frac{1}{\log{p}} = \frac{\pi(n)}{\log n}+\int_2^n\frac{\pi(x)}{x\log^2 x}dx. $$

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I was looking at this just last night. What are the odds? (Someone else can pose that as an MO question.) –  The Masked Avenger Nov 15 '13 at 22:40
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