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Let $a$ be a non-negative integer. Is it true that $$ \prod_{p\leq x} \left(1+\frac{a}{p}\right)\leq \left(\frac{\log{x}}{\log{2}}\right)^a?$$

Edit: If we apply the logarithm function to the product $\prod_{p\leq x} (1+a/p)$ we get $\sum_{p\leq x} \log(1+a/p) .$ Since $\log(1+x)\sim x$ when $x$ tends to $0$ then by Mertens theorem, we obtain $$\sum_{p\leq x} \log(1+a/p)\sim a \sum_{p\leq x} 1/p=a\log{\log{x}}+O(1)=\log{(\log{x})^a}-\log{(\log{2})^a}+O(1).$$ Then, we conclude that in the asymptotic limit ${x \rightarrow \infty}$, we have $$\prod_{p\leq x} (1+a/p)\leq\frac{(\log{x})^a}{(\log{2})^a}.$$ Is it true what I wrote? Thanks in advance.

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  • $\begingroup$ @Crostul if we apply the logarithm function to the product $\prod_{p\leq x} (1+a/p)$ we get $\sum_{p\leq x} \log(1+a/p) .$ Since $\log(1+x)\sim x$ when $x$ tends to $0$ then by Mertens theorem, we obtain $$\sum_{p\leq x} \log(1+a/p)\sim a \sum_{p\leq x} 1/p=a\log{\log{x}}+O(1)=\log{(\log{x})^a}-\log{(\log{2})^a}+O(1).$$ Then, we conclude that in the asymptotic limit ${x \rightarrow \infty}$, we have $$\prod_{p\leq x} (1+a/p)\leq\frac{(\log{x})^a}{(\log{2})^a}.$$ $\endgroup$ – Khadija Mbarki Sep 13 '16 at 21:36
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I don't think it is correct. Note that if you have an asymptotic $$f\left(n\right)=g\left(n\right)+O\left(1\right) $$ it doesn't mean that $$f\left(n\right)\leq g\left(n\right) $$ for example $$\sum_{k\leq n}\frac{1}{k}=\log\left(n\right)+O\left(1\right) $$ but from the integral test we have $$\sum_{k\leq n}\frac{1}{k}\geq\int_{1}^{n+1}\frac{1}{t}dt=\log\left(n+1\right). $$ I think we can proceed as follow. Since $\log\left(1+x\right)<x $ we have $$\sum_{p\leq x}\log\left(1+\frac{a}{p}\right)\leq a\sum_{p\leq x}\frac{1}{p}. $$ Now we know that $$\sum_{p\leq x}\frac{1}{p}=\log\left(\log\left(x\right)\right)+M+O\left(\frac{1}{\log\left(x\right)}\right)\tag{1} $$ where $M$ is the Meissel–Mertens constant. From $\left(1\right) $ we have that exists some $A>0 $ such that $$\left|\sum_{p\leq x}\frac{1}{p}-\log\left(\log\left(x\right)\right)-M\right|\leq\frac{A}{\log\left(x\right)} $$ and if $x$ is sufficiently large we have $$\sum_{p\leq x}\frac{1}{p}\leq\log\left(\log\left(x\right)\right)+M+\frac{A}{\log\left(x\right)}<\log\left(\log\left(x\right)\right)-\log\left(\log\left(2\right)\right) $$ since $M\approx0.2614 $ and $-\log\left(\log\left(2\right)\right)\approx0.3665 $. But this holds only for a sufficiently large $x$. If you want a result for all $x$ we can argue as follows. Again from $(1)$ exists some constant $C>0 $ such that $$\sum_{p\leq x}\frac{1}{p}\leq\log\left(\log\left(x\right)\right)+C $$ for example in this paper (Wayback Machine) is shown that $$\left|\sum_{p\leq x}\frac{1}{p}-\log\left(\log\left(x\right)\right)\right|<6 $$ if $x>e^{4}\approx54.598 $. So if you want a result for all $x$ you have to find a constant that holds also for $x\in\left[2,54\right] .$ Hence we can conclude that $$\prod_{p\leq x}\left(1+\frac{a}{p}\right)\leq\log^{a}\left(x\right)e^{C},\, x\geq2$$

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  • $\begingroup$ From (1) can we say that for sufficiently large $x$, we have $\sum_{p\leq x}\frac{1}{p}>\log\left(\log\left(x\right)\right){\color{\red}-}M$?. How can we deduce a lower bound from an asymptotic formula like (1) ? Thanks $\endgroup$ – mike Sep 15 '16 at 15:56
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    $\begingroup$ @mike For a sufficiently large $x$ we have that $A/\log(x) $ is smaller than every positive constant, and in particular it is smaller than $2M$. So from $(1)$ we have $$\sum_{p\leq x}\frac{1}{p}\geq\log(\log(x))+M-\frac{A}{\log(x)}$$ $$>\log(\log(x))+M-2M=\log(\log(x))-M.$$ $\endgroup$ – Marco Cantarini Sep 15 '16 at 16:31
  • $\begingroup$ Assume that $f(n)=g(n)(1+O(n^{-1}))$ and $g(n)>0$. Is it obvious that $f(n)>({\color{red}{1/2}})g(n)>0$? Does it need a proof? $\endgroup$ – mike Sep 23 '16 at 16:55
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    $\begingroup$ @mike If $f(n)=g(n)(1+O(n^{-1}))$ we have that exists some $C>0$ such that $\left|f(n)-g(n)\right|<Cg(n)/n$ so in particular $$f(n)>g(n)(1-C/n)$$ so for a sufficiently large $n$ your inequality holds. $\endgroup$ – Marco Cantarini Sep 23 '16 at 18:43
  • $\begingroup$ Thanks a lot for your proof! Best $\endgroup$ – mike Sep 23 '16 at 19:51

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