I was wondering whether there exists a known upperbound for:


For example:


I've searched around for a bit, but since english is not my native language, I've been unable to phrase this question in a way that google understands.

I'm really hoping for something in terms of $\log(n)$ or better.

Any kind of help is really appreciated.

  • 1
    $\begingroup$ How about $\sum_{i=2}^{n} \frac {1}{p_i} < f(n)< \frac{n}{3}$ ? $\endgroup$ – rtybase Apr 3 '16 at 21:36
  • $\begingroup$ Yes, that's indeed an upperbound for this function. So it should be a valid answer. But I hope there's also an upperbound in terms of $\log(n)$, I''ll edit my post, since it's a bit unclear $\endgroup$ – Mastrem Apr 3 '16 at 21:39
  • $\begingroup$ The best I can think of then is $\prod_{k=2}^{i} \frac{p_{k}-2}{p_{k}} < \prod_{k=2}^{i} \frac{p_{k}-1}{p_{k}} \sim \frac{e^{-\gamma }}{\ln{n}}$ which is Merten’s theorem (section 22.8 in this book matematica.cubaeduca.cu/medias/pdf/842.pdf). $\endgroup$ – rtybase Apr 3 '16 at 22:03
  • $\begingroup$ That would be a pretty good bound. Do you know whether $\prod_{k=2}^{n}\dfrac{p_k-2}{p_k}<\dfrac{e^{-\gamma}}{ln\ n}$ always holds? $\endgroup$ – Mastrem Apr 3 '16 at 22:07
  • $\begingroup$ Just make sure you deal with the case $k=1$, e.g. $\prod_{k=2}^{n} \frac{p_{k}-1}{p_{k}} = 2 \cdot \prod_{k=1}^{n} \frac{p_{k}-1}{p_{k}} \sim \frac{2 \cdot e^{-\gamma }}{\ln{n}}$ $\endgroup$ – rtybase Apr 3 '16 at 22:16

Well $$\prod_{k=2}^{n} \frac{p_{k}-2}{p_{k}}<\prod_{k=2}^{n} \frac{p_{k}-1}{p_{k}}=2\prod_{k=1}^{n} \frac{p_{k}-1}{p_{k}}\sim \frac{2e^{-\gamma }}{\ln{n}}$$

This means that there $\exists \varepsilon>0$, constant such that $$\prod_{k=2}^{n} \frac{p_{k}-2}{p_{k}} < (1+\varepsilon) \frac{2e^{-\gamma }}{\ln{n}}$$ always.

For more details see Mertens' theorems. Or section 22.8 of this book.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.