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I was wondering whether there exists a known upperbound for:

$$f(n)=\sum_{i=2}^{n}\bigg(\prod_{k=2}^{i}\dfrac{p_k-2}{p_k}\bigg)$$

For example:

$$f(4)=\dfrac{1}{3}+\dfrac{1\cdot3}{3\cdot5}+\dfrac{1\cdot3\cdot5}{3\cdot5\cdot7}+\dfrac{1\cdot3\cdot5\cdot9}{3\cdot5\cdot7\cdot11}$$

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.

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    $\begingroup$ 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
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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.

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