Proving that $6n-1$ divides the numerator of an Alternating Harmonic Number in simplest form

Question

I'm working on a problem from a number theory book where we consider the alternating harmonic series truncated at the term $\frac{1}{4n-1}$. More precisely, we define

$$S_n = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{4n-1}$$

and suppose that, when written in lowest terms,

$S_n = \frac{a}{b}$ with $\gcd(a, b) = 1$

I need to prove that if $6n-1$ is prime, then $6n-1 \mid a$.

I have some ideas that involve rewriting the series in terms of harmonic numbers and working modulo $6n-1$. Some of my ideas are:

1. Rewriting the Series:
   Express $S$ in terms of harmonic numbers. One way to do this is to show that

   $$S_n = H_{4n-1} - H_{2n-1}$$

   where $H_k = \sum_{i=1}^{k} \frac{1}{i}$ is the $k$th harmonic number.

2. Working in $\mathbb{Z}_{6n-1}$:
   Since $6n-1$ is prime and $6n-1 > 4n-1$, none of the denominators $1, 2, \dots, 4n-1$ is divisible by $6n-1$. Therefore, every integer $k$ in that range is invertible modulo $6n-1$.

3. Pairing and Cancellation:
   By considering the series modulo $6n-1$, try to pair terms in such a way that their contributions cancel, showing that

   $$S_n \equiv 0 \pmod{6n-1}.$$

   This would imply that, in the reduced form $\frac{a}{b}$, the numerator $a$ must be divisible by $6n-1$ (since $b$ is coprime to $6n-1$).

I would appreciate any suggestions or corrections to this approach. Specifically, I am looking for a solution without use of calculus, but basic calculus should be acceptable.

I have seen a similar (but different) question on Math StackExchange regarding the divisibility of the numerator of an alternating harmonic series, which considers

$$\sum_{k=1}^{4n} (-1)^{k-1}\frac{1}{k}$$

with $p=6n+1$ prime.

Any help or pointers would be greatly appreciated. Thank you!

Answer 1

Proof that $p$ Divides the Numerator of the Alternating Harmonic Sum

Theorem.
Let $$ S_n = \sum_{k=1}^n \frac{(-1)^{k-1}}{k} = \frac{a}{b} $$ where $\gcd(a,b)=1$ and $n>1$. If $$ p = n+1+\left\lfloor \frac{n}{2} \right\rfloor $$ is prime, then $p \mid a$.

Proof.
We proceed by expressing $S_n$ in terms of harmonic numbers and analyzing the sum modulo $p$.

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1. Expressing $S_n$ as a Difference of Harmonic Numbers

The alternating harmonic sum can be written as $$ S_n = \sum_{k=1}^n \frac{(-1)^{k-1}}{k} = H_n - H_{\left\lfloor \frac{n}{2} \right\rfloor}, $$ where the $n$th harmonic number is defined by $$ H_n = \sum_{k=1}^n \frac{1}{k}. $$

This follows from writing $$ S_n = \sum_{\substack{k=1 \\ k \text{ odd}}}^n \frac{1}{k} - \sum_{\substack{k=1 \\ k \text{ even}}}^n \frac{1}{k} = H_n - 2\sum_{k=1}^{\left\lfloor \frac{n}{2} \right\rfloor} \frac{1}{2k} = H_n - H_{\left\lfloor \frac{n}{2} \right\rfloor}. $$

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2. Defining $p$ Based on the Parity of $n$

The prime $p$ is defined as $$ p = n + 1 + \left\lfloor \frac{n}{2} \right\rfloor. $$

We consider two cases based on whether $n$ is even or odd:

• Case 1: $n$ is even.
  Let $n = 2m$. Then, $$ \left\lfloor \frac{n}{2} \right\rfloor = m, \quad p = 2m + 1 + m = 3m + 1. $$ The sum becomes $$ S_n = H_{2m} - H_m = \sum_{k=m+1}^{2m} \frac{1}{k}. $$

• Case 2: $n$ is odd.
  Let $n = 2m + 1$. Then, $$ \left\lfloor \frac{n}{2} \right\rfloor = m, \quad p = (2m+1) + 1 + m = 3m + 2. $$ The sum becomes $$ S_n = H_{2m+1} - H_m = \sum_{k=m+1}^{2m+1} \frac{1}{k}. $$

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3. Analyzing the Sum Modulo $p$

We now show that $S_n \equiv 0 \pmod{p}$ by pairing terms in the sum such that each pair sums to a multiple of $p$.

• Case 1: $n = 2m$.
  The sum $$ S_n = \sum_{k=m+1}^{2m} \frac{1}{k} $$ has $m$ terms. Pair the terms as follows: $$ \frac{1}{m+1} + \frac{1}{2m} = \frac{3m+1}{(m+1)(2m)} = \frac{p}{(m+1)(2m)}, $$ $$ \frac{1}{m+2} + \frac{1}{2m-1} = \frac{3m+1}{(m+2)(2m-1)} = \frac{p}{(m+2)(2m-1)}, $$ and so on. In each pair the numerator is $p$, and since $p$ is prime and $p > 2m$, it does not divide any denominator. Therefore, each paired sum is congruent to $0$ modulo $p$, so the entire sum satisfies $$ S_n \equiv 0 \pmod{p}. $$

• Case 2: $n = 2m+1$.
  The sum $$ S_n = \sum_{k=m+1}^{2m+1} \frac{1}{k} $$ has $m+1$ terms. Pair the terms similarly: $$ \frac{1}{m+1} + \frac{1}{2m+1} = \frac{3m+2}{(m+1)(2m+1)} = \frac{p}{(m+1)(2m+1)}, $$ $$ \frac{1}{m+2} + \frac{1}{2m} = \frac{3m+2}{(m+2)(2m)} = \frac{p}{(m+2)(2m)}, $$ and so on. Again, each numerator is $p$, and since $p$ does not divide any denominator, the entire sum is $$ S_n \equiv 0 \pmod{p}. $$

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4. Why $p$ Must Divide the Numerator $a$

Since $$ S_n \equiv 0 \pmod{p}, $$ and we have written $$ S_n = \frac{a}{b} $$ with $\gcd(a,b)=1$, it follows that $$ \frac{a}{b} \equiv 0 \pmod{p}. $$ Because $p$ does not divide $b$ (note that $b$ is composed of factors less than or equal to $n$, and $p>n$), multiplying both sides by $b$ gives $$ a \equiv 0 \pmod{p}. $$ Thus, $p$ divides $a$, which completes the proof.