Harmonic Number(2)

March 2025
M	T	W	T	F	S	S
	1	2
3	4	5	6	7	8	9
10	11	12	13	14	15	16
17	18	19	20	21	22	23
24	25	26	27	28	29	30
31

Our goal is to prove the following theorem by Carlo Sanna. The proof is adapted and changed slightly from his paper.

Theorem. For any prime number $p$ and any $x \ge 1$ , we have

$J_p(x) < 129p^{2/3}x^{0.765}$

Lemma 1 If $p > 3$ is prime, $n\geq 1$ , and $0\leq k\leq p - 1$ , then

$H_{pn}=\frac{1}{p}H_n+O(p^2)$

and

$H_{pn+k} = H_{pn} + H_k + O(p)$

Proof. $H_{pn}-\frac{1}{p}H_n=\sum_{k=0}^{n-1}\sum_{j=1}^{p-1}\frac{1}{kp+j}$ . Each of the sum is $O(p^2)$ by Wolstenholme’s theorem. The second equation holds since

$\frac{1}{i}-\frac{1}{np+i}=\frac{pn}{i(np+i)}=O(p)$

Lemma 2 If $p$ is an odd prime, $n \geq 1$ and $0 \leq k \leq p-1$ , then $v_p(H_{pn+k}) > 0$ if and only if $v_p(H_n) > 0$ and $H_{pn} = -H_k + O(p)$ .

Proof.

From Lemma 1, we have

$H_{pn}=H_{pn+k}-H_{k}+O(p)$

So we need to prove that $v_p(H_k)\leq 0$ , then $v_p(H_{pn}\geq 0$ , giving us $v_p(H_n)>0$ from Lemma 1. In fact $v_p(H_k)=0$ , since $\frac{1}{i}+\frac{1}{np+i}=O(p)$ , and $p$ is odd, so $v_(H_k)\leq 0$ . We also have $v(H_k)\geq 0$ considering the denominators.

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