# An upper bound for $\sqrt{p_{n+1}}$

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)

• 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). Aug 5, 2019 at 14:46
• @alpoge ...and this proves that the inequality does not hold. Aug 6, 2019 at 7:05
• 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! Aug 6, 2019 at 7:17
• @alpoge: Your comment is completely fine except that your sum is asymptotic to twice of what you indicate. Sep 5, 2019 at 17:19
• @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! Sep 5, 2019 at 17:20

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 $$M>4C^2.$$
• 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 $(C+o(1))\sqrt{n\log n}$. Sep 5, 2019 at 16:13