Vtyjkyuk

I am working on an alternative argument for the Sylvester-Schur theorem.

I want to show that for $n > 18$, $n+1 > sqrt[leftroot-1uproot2scriptstyle left(n - pi(n) - 1right)] n! $.

Can be AM-GM be used to demonstrate this?

Here's my thinking:

(1) For $n ge 2, pi(n) le fracn3 + 2$

The prime counting function $pi(n) le fracn3 + 2$ since for a prime $p > 3, p equiv 1 pmod 6$ or $p equiv 5 pmod 6$ so that $pi(n) le 2 + leftlfloorfracn+16rightrfloor + leftlfloorfracn+56rightrfloor - 1 le frac2n+126 = fracn+63$.

(2) For $n > 18, fracn-22 > pi(n)$

If $n > 18$, then $fracn-22 > fracn3+2 ge pi(n)$ since $3n - 6 > 2n+12$.

(3) $1 > dfracn2(n-pi(n) -1)$

This follows from $2(n - pi(n) -1) > n$ since $n-2 > 2pi(n)$.

(4) Multiplying $n+1$ to both sides of the equation:

$$n+1 > fracn(n+1)2(n-pi(n)-1)$$

Can I now use AM-GM and the summation formula of $sumlimits_i le ni = frac(n+1)(n)2$ to conclude:

$$n+1 > fracn(n+1)2(n - pi(n)-1) > sqrt[leftroot-1uproot2scriptstyle left(n-pi(n)-1right)]n!$$

Strictly speaking, $sqrt[k]n! not leq fracn(n+1)2k$ for all $k geq 1$. (The Left Hand Side converges to 1 and the right converges to 0 as $k to infty.)

The inequality fails for $n = 10$, where $pi(10) = 4$ and so $$frac10(11)2(10-4-1) = 11 < 20.5093835306 approx sqrt[10-4-1]10!$$ and for $n = 19$, $pi(19) = 8$, so $$frac19(20)2(19-8-1) = 19 < 51.1104214517 approx sqrt[10]19!$$

Even for $n = 100$, $pi(100) = 25$ (pi(n)) so
$$frac100(101)2(100-25-1) = 68.2432432432 < 136.373434808 approx sqrt[10-25-1]100!$$

So your inequality fails for $n$ as large as $100$.(You can check my work, if you want.)

The only redeeming property seems to be that for $k_n = n-pi(n) -1 < n$, $fracCnlog(n)leq pi(n) leq fracDnlog(n)$ so $k_n$ is slightly smaller than $1/n$. See (1) below. Even without that, $k_n > 0$, as you showed above.

A direct approach does not seem to work (but there might be others), because:

$$(n!)^1/k_n = (n!)^1/n(n!)^1/k_n - 1/n < fracn(n+1)2n(n!)^(pi(n)+1)/(n(n-pi(n)-1)) = fracn(n+1)2(n-pi(n)-1) left(fracn-pi(n)-1nright)(n!)^(pi(n)+1)/(n(n-pi(n)-1)) < fracn(n+1)2(n-pi(n)-1) (n!)^(pi(n)+1)/(n(n-pi(n)-1)).$$

The question is how to control that exponent. which you need to be less than $1/n$, because you need the factorial term to vanish. If you use the estimates for $pi(n)$, then you can get that it is at most
$$fracDn/log(n)+1n(n-Cn/log(n)-1) = fracD/log(n)+1/nn(1-C/log(n)-1/n)$$
which is at most $Const.left(frac1nlog(n)right)$. But since $sqrt[n]n! sim n/e$ by Stirling's Inequality and $n^1/log(n) = e$, (See here) we see that this term is just bounded by a constant. So, at the end of the day what I got for large $n$ by applying AM-GM in "the obvious way" was $(n!)^1/k_n < Const. fracn(n+1)2(n-pi(n)-1)$

(1) $C, D > 0$ exist (perhaps for large $n$) according to an "elementary" proof (meaning just combinatorics, not complex analysis or such) in Number Theory - Andrews, (I don't have the book with me) and $C = 1, D = 1.25506$ according to Prime Counting