[Paper Review] Congruences involving alternating multiple harmonic sum

Roberto Tauraso|ArXiv.org|May 20, 2009

Advanced Mathematical Identities7 references4 citations

TL;DR

This paper establishes a p-adic congruence for an alternating multiple harmonic sum involving the binomial coefficient $\binom{-1/2}{k}$, showing it is congruent modulo $p^3$ to the negative harmonic sum $-\sum_{k=1}^{(p-1)/2} \frac{1}{k}$ for primes $p > 3$. The result refines a known $p$-adic logarithmic identity using Fermat quotient and Bernoulli number terms, derived via advanced congruences on alternating multiple harmonic sums and binomial coefficient expansions.

ABSTRACT

We show that for any prime prime $p ot=2$ $$\sum_{k=1}^{p-1} {(-1)^k\over k}{-{1\over 2} \choose k} \equiv -\sum_{k=1}^{(p-1)/2}{1\over k} \pmod{p^3}$$ by expressing the l.h.s. as a combination of alternating multiple harmonic sums.

Motivation & Objective

• To establish a $p^3$-congruence for the sum $\sum_{k=1}^{p-1} \frac{(-1)^k}{k} \binom{-1/2}{k}$ using $p$-adic analysis.
• To refine the known $p$-adic logarithmic identity $\sum_{k=1}^{p-1} \frac{(-1)^k}{k} \binom{-1/2}{k} \equiv 2q_p(2) \pmod{p}$ to higher precision.
• To derive new congruences for alternating multiple harmonic sums of depth 2 and negative indices modulo $p$.
• To connect the sum to the harmonic sum $-\sum_{k=1}^{(p-1)/2} \frac{1}{k}$ via $p^3$-congruences involving $q_p(2)$ and $B_{p-3}$.

Proposed method

• Expresses the sum $\sum_{k=1}^{p-1} \frac{(-1)^k}{k} \binom{-1/2}{k}$ as a combination of alternating multiple harmonic sums using generating functions.
• Applies the reversal and shuffle relations for alternating multiple harmonic sums to reduce depth-2 sums modulo $p$.
• Uses known congruences for non-alternating harmonic sums modulo $p^2$ and $p^3$, including those involving Bernoulli numbers and Fermat quotients.
• Employs binomial coefficient expansions modulo $p^3$, particularly $\binom{2p}{j} \equiv -2p/j + 4p^2 H(1;j-1)/j \pmod{p^3}$.
• Applies the identity $4^{p-1} \equiv 1 - 2q_p(2)p + 3q_p(2)^2p^2 \pmod{p^3}$ to invert the scaling factor in the sum.
• Combines all terms via algebraic manipulation and modular inversion to derive the final $p^3$-congruence.

Experimental results

Research questions

• RQ1 What is the $p^3$-adic refinement of the known $p$-adic congruence $\sum_{k=1}^{p-1} \frac{(-1)^k}{k} \binom{-1/2}{k} \equiv 2q_p(2) \pmod{p}$?
• RQ2 How can alternating multiple harmonic sums of negative indices be related to classical harmonic sums and Bernoulli numbers modulo $p^3$?
• RQ3 Can the sum $\sum_{k=1}^{p-1} \frac{(-1)^k}{k} \binom{-1/2}{k}$ be expressed as a $p^3$-congruence to $-\sum_{k=1}^{(p-1)/2} \frac{1}{k}$?
• RQ4 What new congruences emerge for alternating multiple harmonic sums such as $H(-1,-2;p-1)$ and $H(-1,-1,1;p-1)$ modulo $p$?

Key findings

• The sum $\sum_{k=1}^{p-1} \frac{(-1)^k}{k} \binom{-1/2}{k}$ is congruent to $2q_p(2) - p q_p(2)^2 + \frac{2}{3}p^2 q_p(2)^3 + \frac{7}{12}p^2 B_{p-3} \pmod{p^3}$ for primes $p > 3$.
• This sum is congruent to $-\sum_{k=1}^{(p-1)/2} \frac{1}{k} \pmod{p^3}$, linking it directly to a classical harmonic sum.
• The congruence $H(-1,-2;p-1) \equiv -\frac{3}{4}B_{p-3} \pmod{p}$ holds for primes $p \neq 2$.
• The congruence $H(-1,-1,1;p-1) \equiv q_p(2)^3 + \frac{7}{8}B_{p-3} \pmod{p}$ is established for primes $p \neq 2$.
• The identity $4^{p-1} \sum_{k=1}^{p-1} \frac{(-1)^k}{k} \binom{-1/2}{k} \equiv 2q_p(2) + 3p q_p(2)^2 + \frac{2}{3}p^2 q_p(2)^3 + \frac{7}{12}p^2 B_{p-3} \pmod{p^3}$ is derived via binomial coefficient expansions.
• The inverse scaling $4^{-(p-1)} \equiv 1 - 2q_p(2)p + 3q_p(2)^2p^2 \pmod{p^3}$ is used to recover the original sum modulo $p^3$.