0
$\begingroup$

Let $C$ be a positive constant. Is it true that for all sufficiently large integers $n$ the inequality $$\prod_{i=1}^n (1+\frac{1}{\sqrt{p_i}})>C\sqrt{p_{n+1}}$$ holds? (Here with $p_k$ is denoted the $k-$th prime number)

$\endgroup$
5
  • 6
    $\begingroup$ Yep take logs and insert \sum_{p\leq X} 1/\sqrt{p} ~ X^{1/2} / \log{X} (by partial summation) after using \log(1+x) = x + O(x^2). $\endgroup$
    – alpoge
    Aug 5 '19 at 14:46
  • $\begingroup$ @alpoge ...and this proves that the inequality does not hold. $\endgroup$ Aug 6 '19 at 7:05
  • 4
    $\begingroup$ I disagree but may be being stupid! I think the lower bound gotten is exp of \sqrt{p_n} / \log{p_n} or some such! $\endgroup$
    – alpoge
    Aug 6 '19 at 7:17
  • $\begingroup$ @alpoge: Your comment is completely fine except that your sum is asymptotic to twice of what you indicate. $\endgroup$
    – GH from MO
    Sep 5 '19 at 17:19
  • $\begingroup$ @GHfromMO Indeed I should know not to write $\sim$ when ignoring constants. One has to divide by the exponent 1/2 when integrating and thus the factor of two. Sorry! $\endgroup$
    – alpoge
    Sep 5 '19 at 17:20
1
$\begingroup$

In fact, it is true. First of all denote the expression on the LHS of the inequality with $X.$ For now, fix $n$(which we will choose large enough later). Let $P$ be the set of the first $n$ primes. For a positive integer $N$ we denote by $f(N)$ the number of integers less than or equal to $N$ with squarefree parts, belonging to $P.$ Now let $A$ be a subset of $P$ and let $P(A)$ denotes the product of the elements of $A.$ The number of integers of the form $m^2A$ less than $N$ is exactly $[{\sqrt{\frac{N}{P(A)}}}],$ which is not greater than $\sqrt{\frac{N}{P(A)}}.$ Thus, $$f(N)\leq \sum_{A\subset P} \sqrt{\frac{N}{P(A)}}.$$ Note that the RHS of the inequality is exactly $$\sqrt{N}\prod_{i=1}^{n} (1+\frac{1}{\sqrt{p_i}}),$$ therefore $$X\geq \frac{f(N)}{\sqrt{N}}.$$ Now, let $M$ be a large enough integer and let $N=Mp_{n+1}.$ Let $p_{n+2},\ldots,p_{n+s}$ be all the primes greater than $p_{n+1}$ and smaller than $Mp_{n+1}.$ For each $M$ we can choose $n$ to be large enough such that $p_{n+1}>M.$ Now the numbers in $[1,N],$ which aren't counted in $f(N)$ are those divisible by $p_{n+i}(i=1,\ldots,s).$ So, $$f(N)\geq N(1-\sum_{i=1}^s \frac{1}{p_{n+i}}).$$ Now, pick $n,$ such that $\log{n}>10M.$ It is well known (around the prime number theorem) that $p_n>\frac{n\log{n}}{5}$ for each $n,$ so $Mp_{n+1}>p_{n+s}>\frac{(n+s)\log{(n+s)}}{5}.$ Hence, $$n+s<\frac{5Mp_{n+1}}{\log{(n+s)}}<\frac{5Mp_{n+1}}{\log{n}}<\frac{p_{n+1}}{2}.$$ As a consequence, $$\sum_{i=1}^{s} \frac{1}{p_{n+i}}<\frac{s}{p_{n+1}}<\frac{1}{2}.$$ Therefore, $$X\geq\frac{f(Mp_{n+1})}{\sqrt{Mp_{n+1}}}\geq \frac{Mp_{n+1}}{2\sqrt{Mp_{n+1}}}\geq \frac{\sqrt{Mp_{n+1}}}{2}>C\sqrt{p_{n+1}}$$ for $M>4C^2.$

$\endgroup$
1
  • 1
    $\begingroup$ User alpoge has already explained, in a comment below the original post, that the left hand side is much larger than the right hand side. Well, he is missing a constant, the left hand side is about the square of what he claims. Precisely, the left hand side is $\exp((2+o(1))\sqrt{n/\log n})$, while the right hand side is $(C+o(1))\sqrt{n\log n}$. $\endgroup$
    – GH from MO
    Sep 5 '19 at 16:13

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.